qid int64 1 4.65M | question large_stringlengths 27 36.3k | author large_stringlengths 3 36 | author_id int64 -1 1.16M | answer large_stringlengths 18 63k |
|---|---|---|---|---|
1,510 | <p>It's hard to prove a number is transcendental (non-algebraic) yet there are some wonderful examples amongst them like π,e and Liouville's number. What's so special about them? </p>
<p>Are most numbers transcendental?</p>
| Michael Lugo | 143 | <p>Most numbers are transcendental. In particular, the set of algebraic numbers is countable -- basically because there are countably many polynomials, each with countably many roots. (This is a generalization of the argument that the rationals are countable.) But there are uncountably many real numbers.</p>
|
1,450,176 | <p>I would like to evaluate this limit :$$\displaystyle \lim_{x \to \infty} ({x\sin \frac{1}{x} })^{1-x}$$.</p>
<p>I used taylor expansion at $y=0$ , where $x$ go to $\infty$ i accrossed this </p>
<p>problem : ${1}^{-\infty }$ then i can't judge if this limit equal's $1$ , </p>
<p>because it is indeterminate case ,Then is there a mathematical way to </p>
<p>evaluate this limit ?</p>
<p>Thank you for any help </p>
| Ron Gordon | 53,268 | <p>I think you have to expand out to the next order term in the Taylor series about infinity. Thus</p>
<p>$$\left ( x \sin{\frac1{x}} \right )^{1-x} = \frac{\displaystyle 1-\frac1{6 x^2}}{\displaystyle\left (1-\frac1{6 x^2} \right )^x}$$</p>
<p>Now</p>
<p>$$\left (1-\frac1{6 x^2} \right )^x = \left[ \left (1-\frac1{6 x^2} \right )^{x^2}\right ]^{1/x}$$</p>
<p>so as $x \to \infty$, the expression approaches</p>
<p>$$ \lim_{x \to \infty}\frac{\displaystyle 1-\frac1{6 x^2}}{\displaystyle e^{-1/(6 x)}} = 1$$</p>
|
4,620,310 | <p>Prove <span class="math-container">$\lim_{x\to\infty}\sin x/x=0$</span> using definition of a limit of a function.</p>
<p>I know that the definition of a limit of a fuction <span class="math-container">$f(x)$</span> (when <span class="math-container">$x\to\infty$</span> and <span class="math-container">$f(x)$</span> has a finite limit) as: <span class="math-container">$\forall \epsilon >0$</span>, <span class="math-container">$\exists M>0$</span> such that <span class="math-container">$x> M$</span>, <span class="math-container">$|f(x)-L|<\epsilon .$</span> But, I dont know how to find <span class="math-container">$M$</span> (in terms of <span class="math-container">$\epsilon $</span>) in order to prove <span class="math-container">$|\frac {\sin x}{ x}|<\epsilon$</span> ? Also we need to prove that using this definition, I stated above.</p>
| Lorago | 883,088 | <p>You're a bit wrong with what you need to prove. In particular, since you want the limit to be <span class="math-container">$0$</span>, you want <span class="math-container">$\left\lvert\frac{\sin x}{x}\right\rvert<\varepsilon$</span> and <strong>not</strong> <span class="math-container">$\left\lvert\frac{\sin x}{x}-1\right\rvert<\varepsilon$</span>, which is what you would consider if you wanted the limit to be <span class="math-container">$1$</span>.</p>
<p>Now let <span class="math-container">$\varepsilon>0$</span>. Notice that, since <span class="math-container">$\lvert\sin x\rvert\leq 1$</span> for all <span class="math-container">$x\in\mathbb{R}$</span> we have that</p>
<p><span class="math-container">$$\left\lvert\frac{\sin x}{x}\right\rvert\leq\frac{1}{\lvert x\rvert}$$</span></p>
<p>for all <span class="math-container">$x\in\mathbb{R}\setminus\{0\}$</span>. Now suppose we have an <span class="math-container">$M>0$</span> and let <span class="math-container">$x>M$</span>. Then the above inequality gives us that</p>
<p><span class="math-container">$$\left\lvert\frac{\sin x}{x}\right\rvert<\frac{1}{M},$$</span></p>
<p>and so in particular if we choose <span class="math-container">$M=\frac{1}{\varepsilon}$</span>, it follows from the above that</p>
<p><span class="math-container">$$\left\lvert\frac{\sin x}{x}\right\rvert<\varepsilon$$</span></p>
<p>for all <span class="math-container">$x>M$</span>, proving that</p>
<p><span class="math-container">$$\lim_{x\to\infty}\frac{\sin x}{x}=0.$$</span></p>
|
3,867,197 | <p>Let <span class="math-container">$A$</span> be the following matrix</p>
<p><span class="math-container">$$\left(
\begin{array}{ccc}
1 & 0 & x \\
0 & 1 & y \\
x & y & 1
\end{array}
\right)$$</span></p>
<p>I have to prove that if, at least <span class="math-container">$x+y>\frac{3}{2}$</span>, <span class="math-container">$A$</span> is not positive definite.</p>
<p>I have tried to prove it by calculating the eigenvalues, obtaining:
<span class="math-container">$$
\begin{array}{c}
\lambda_1=1\\
\lambda_2=1+\sqrt{x^2+y^2} \\
\lambda_3=1-\sqrt{x^2+y^2}
\end{array}
$$</span></p>
<p>It is obvious that <span class="math-container">$\lambda_1$</span> and <span class="math-container">$\lambda_2$</span> are always positive, so I only have to take care of <span class="math-container">$\lambda_3$</span>. The problem is that I cannot relate the given condition with <span class="math-container">$1-\sqrt{x^2+y^2}<0$</span>, which would prove that the matrix is not positive definite.</p>
| player3236 | 435,724 | <p>For <span class="math-container">$x+y>\frac32$</span>:</p>
<p><span class="math-container">$$x^2+y^2=\frac12(x^2+y^2+x^2+y^2)\ge\frac12(x^2+y^2+2xy) =\frac12(x+y)^2>\frac98>1$$</span></p>
|
1,879,673 | <p>I have woven the below incomplete proof of the following claim:</p>
<blockquote>
<p><em>Claim</em>. If $X$ is completely regular and $Y$ is a compactification of $X$,
then there is a unique, continuous, surjective, closed map
$g:\beta\left(X\right)\to Y$ which is the identity on
$X$.</p>
</blockquote>
<p><em>Here, $\beta\left(X\right)$ is the Stone-Čech compactification of $X$.</em></p>
<p><em>Proof</em>. Let $f:X\to Y$ be such that $x\mapsto x$. Then $f$ is continuous. Since $f$ is continuous and $Y$ is compact and Hausdorff, it is the case that $f$ extends uniquely to a continuous map $g:\beta\left(X\right)\to Y$. Then $g$ is the identity on $X$. Let $C\subseteq\beta\left(X\right)$ be closed. Since $\beta\left(X\right)$ is compact, it is the case that $C$ is compact. Since $g$ is continuous, it is the case that $g\left(C\right)$ is compact. Since $Y$ is Hausdorff, it is the case that $g\left(C\right)$ is closed. Then $g$ is closed...</p>
<p>I do not know how to show that $g$ is surjective. Am I allowed to use the "maximality" of $\beta\left(X\right)$? If so, then I believe that it would follow that $Y\subseteq\beta\left(X\right)$, which would imply that $g$ is surjective. I am not sure because this "maximality" is not defined in terms of containment.</p>
| Brian M. Scott | 12,042 | <p>Suppose that $f[X]\subsetneqq Y$, let $U=Y\setminus f[\beta X]$, and fix $y\in U$. Then $U$ is an open nbhd of $y$ in $Y$, and $f[X]$ is dense in $Y$, so $U\cap f[X]\ne\varnothing$. Do you see the contradiction?</p>
<p>You’re right to worry about assuming that $Y\subseteq\beta X$: it need not be true, because — as you say — <em>maximality</em> has a different meaning here.</p>
|
4,214,329 | <p>For convenience, let <span class="math-container">$(f(x), g(x))$</span> be a solution to the problem. Now,
<span class="math-container">\begin{align*}
f(x) + g(x) &= f(x)g(x) \\
f(x)g(x) - f(x) - g(x) &= 0 \\
f(x)g(x) - f(x) - g(x) + 1 &= 1 \\
(f(x) - 1)(g(x) - 1) &= 1
\end{align*}</span></p>
<p>By letting <span class="math-container">$f(x) - 1 = 1$</span> and <span class="math-container">$g(x) - 1 = 1$</span>, we get the solution <span class="math-container">$(2,2)$</span>. Also, letting <span class="math-container">$f(x) - 1 = -1$</span> and <span class="math-container">$g(x) - 1 = -1$</span>, we get the solution <span class="math-container">$(0,0)$</span>.</p>
<p>What I am wondering now, is if there exists <span class="math-container">$f$</span> and <span class="math-container">$g$</span> where both are nonconstant polynomials?</p>
<hr />
<p>Edit 1. The set of real numbers is the domain and range of both <span class="math-container">$f$</span> and <span class="math-container">$g$</span>.</p>
| YiFan | 496,634 | <p>You showed it yourself! The only way the product of two polynomials can be constant is if they are themselves constant: otherwise, the degree of the product polynomial would not be zero. Thus <span class="math-container">$(f(x)-1)(g(x)-1)=1$</span> immediately implies <span class="math-container">$f,g$</span> are constant.</p>
|
145,950 | <p>how can I show that any finite CW-space can embedded into an euclidean space of some dimension? Any help or reference would be greatly appreciated.</p>
| Wlodek Kuperberg | 36,904 | <p>If your finite CW-complex is of topological dimension $n$, then it is an $n$-dimensional compact metric space, thus, by the The Menger-Nöbeling theorem (1932), it can be embedded in ${\mathbb R}^{2n+1}$. In this theorem $2n+1$ is the lowest possible dimension, since there exist $n$-dimensional simplicial complexes that cannot be embedded in ${\mathbb R}^{2n}$.</p>
|
15,235 | <p>More precisely, is there a map of schemes $X$ --> $Y$ such that $f$ gives a homeomorphism between $X$ and a closed subset of $Y$, but the corresponding map on sheaves is not surjective?</p>
| Emerton | 2,874 | <p>Yes, for example if $K \subset L$ is an inclusion fields, then the induced map
Spec $L \to $ Spec $K$ is a homeomorphism (both source and target are single points),
but the induced map on sheaves is the given inclusion of $K$ into $L$, which is
surjective only if $K = L$.</p>
<p>For another example, let $X'\to Y$ be a closed immersion of schemes over ${\bar{\mathbb F}}\_p$, and let $X \to X'$ be the relative Frobenius morphism.
Then $X\to X'$ is a homeomorphism on underlying topological spaces but is not an isomorphism of schemes, and so the composite $X\to Y$ is a closed embedding on underlying spaces but not a closed immersion of schemes.</p>
<p>As one last example, let $X'$ be the cuspidal cubic given by $y^2 = x^3$ in the affine
plane $Y$ (over $\mathbb C$, say), and let $X$ be the normalization of $X'$ (which is just
the affine line). Then $X \to X'$ is a homeomorphism on underlying spaces, but is not
an isomorphism of schemes. The composite $X \to Y$ is thus not a closed immersion,
but induces a closed embedding of underlying topological spaces.</p>
|
699,383 | <p>I am a non-mathematician who knows some elemententary calculus ans I want to prove that the sequence $(x_n)$ given by</p>
<p>$$
x_n=-\sqrt{n} + n\ln\Big(1+\frac{1}{\sqrt{n}}\Big)
$$</p>
<p>is decreasing. Is there an elegant way to show this?</p>
| Albert | 82,854 | <p>The simplest way might be to expand (n+1)^(1/2) and (n+1)^(-1/2) in the expression for x_(n+1) using the binomial theorem then show that x_(n+1) - x_n is positive.</p>
|
3,753,819 | <p><span class="math-container">$\textbf{Question:}$</span> Let <span class="math-container">$(X,\mathcal{F},\mu)$</span> be an arbitrary measure space. Let <span class="math-container">$\varphi: \mathbb{R} \rightarrow \mathbb{R}$</span> be continuous and satisfy for some <span class="math-container">$K>0$</span>:</p>
<p><span class="math-container">$$ \vert \varphi(t) \vert \leq K \vert t \vert, \forall t\in \mathbb{R} (*)$$</span></p>
<p>If <span class="math-container">$f \in L^p$</span>, then <span class="math-container">$\varphi \circ f$</span> belongs to <span class="math-container">$L^p$</span>. Conversely, if <span class="math-container">$\varphi$</span> does not satisfy (*), there exists a measure space <span class="math-container">$(X,\mathcal{F},\mu)$</span> and a function <span class="math-container">$f \in L^p$</span> such that <span class="math-container">$\varphi \circ f$</span> does not belong to <span class="math-container">$L^p$</span>.</p>
<p><span class="math-container">$\textbf{My attempt:}$</span> If <span class="math-container">$\varphi$</span> satisfy <span class="math-container">$(*)$</span> we have for each <span class="math-container">$(X,\mathcal{F},\mu)$</span> and <span class="math-container">$x\in X$</span></p>
<p><span class="math-container">$$ \vert (\varphi \circ f)(x) \vert = \vert \varphi(f(x)) \vert \leq K \vert f(x) \vert $$</span></p>
<p><span class="math-container">$$ \implies \vert \varphi \circ f \vert^p \leq K^p \vert f \vert^p $$</span></p>
<p>So <span class="math-container">$\varphi \circ f \in L^p$</span>. I can't solve the other statement, help please.</p>
| Kavi Rama Murthy | 142,385 | <p>The result is not true as stated. In the inequality for <span class="math-container">$\phi$</span> we should add 'for <span class="math-container">$|t|$</span> sufficiently large' to make it correct.</p>
<p>First a counterexample: suppose the measure space is finite and <span class="math-container">$\phi (t)=\sqrt {|t|}$</span>. Then the inequality fails but <span class="math-container">$\phi(f) \in L^{p}$</span> whenever <span class="math-container">$f \in L^{p}$</span>.</p>
<p>Here is the proof when the question is modified as stated:</p>
<p>If (*) is false after the modification then there exist <span class="math-container">$t_n$</span> increasing t0 <span class="math-container">$\infty$</span> such that <span class="math-container">$\phi (t_n) >nt_n$</span>. Consider the real line with Lebesgue measure and define <span class="math-container">$f(x)$</span> to be <span class="math-container">$t_n$</span> on <span class="math-container">$(t_n-r_n, t_n+r_n)$</span> where <span class="math-container">$r_n$</span>'s are so small that the intervals <span class="math-container">$(t_n-r_n,t_n+r_n)$</span> are disjoint. Take <span class="math-container">$f$</span> to be <span class="math-container">$0$</span> outside these intervals. Then <span class="math-container">$\int|f|^{p}=2\sum t_n^{p} r_n$</span> and <span class="math-container">$\int|\phi(f)|^{p}>2\sum n^{p}t_n^{p} r_n$</span>. We only have to choose <span class="math-container">$r_n=\frac 1{n^{p}t_n^{p+1}}$</span>.[ You can always choose <span class="math-container">$t_n$</span> to tend to <span class="math-container">$\infty$</span> as fast as you want so the disjointness condition can be met easily].</p>
|
3,079,023 | <p>A vector space with norm <span class="math-container">$\parallel\cdot\parallel$</span> Satisfy for two vectors the following</p>
<p><span class="math-container">$\parallel x+y\parallel=\parallel x\parallel +\parallel y\parallel$</span></p>
<p>i need to proof the fallowing statement</p>
<p><span class="math-container">$\parallel \alpha x+\beta y\parallel=\alpha\parallel x\parallel +\beta\parallel y\parallel$</span></p>
<p>for all <span class="math-container">$\alpha\geq0$</span> and <span class="math-container">$\beta\geq0$</span>, and also this norm it's not necessary induced by an inner product</p>
<p>The fact that </p>
<p><span class="math-container">$\alpha\parallel x\parallel +\beta\parallel y\parallel\geq\parallel \alpha x + \beta y\parallel$</span></p>
<p>it's clear because the property of the norm, what i can't proof it's the fact that </p>
<p><span class="math-container">$\alpha\parallel x\parallel +\beta\parallel y\parallel\leq\parallel \alpha x + \beta y\parallel$</span></p>
<p>i will appreciate any help.</p>
| Mustafa Said | 90,927 | <p>To show that the norm is induced by an inner product you can show that the norm satisfies the parallelogram law and then use the polarization identity to recover the norm.</p>
|
3,248,569 | <p>I have the following two parametric equations of lines:</p>
<p><span class="math-container">$$\begin{cases} x = -t + 1 \\ y = t + 3 \\ z = -6t \end{cases} \quad \land \quad \begin{cases} x = 2s + 4 \\ y = -s \\ z = 2s + 1 \end{cases}$$</span></p>
<p>I want to examinate their mutual position, that is, I want to find out if they intersect.</p>
<p><span class="math-container">$$\begin{cases} -t + 1 = 2s + 4 \\ t + 3 = -s \\ -6t = 2s + 1 \end{cases}$$</span></p>
<p><span class="math-container">$$\begin{cases} -t + 1 = 2s + 4 \\ -t - 3 = s \\ -6t = 2(-t-3) + 1 \Rightarrow -6t = -2t - 5 \Longrightarrow t = \frac{5}{4}\end{cases}$$</span></p>
<p><span class="math-container">$$-\frac{5}{4} - 3 = s \Longrightarrow s = - \frac{17}{4}$$</span></p>
<p>So they intersect for <span class="math-container">$t = \frac{5}{4} \lor s=-\frac{17}{4}$</span>.</p>
<p>Plugging <span class="math-container">$t$</span> into first equation:</p>
<p><span class="math-container">$$\begin{cases} x = -\frac{5}{4} + 1 \\ y = \frac{5}{4} + 3 \\ z = -6 * \frac{5}{4} \end{cases}$$</span></p>
<p>Point <span class="math-container">$P(-\frac{1}{4}, \frac{17}{4}, -\frac{15}{2})$</span> is the intersection point. </p>
<p>Thing is, I don't think that's true. I sketched those lines in parametric plotter and they look like this:</p>
<p><a href="https://i.stack.imgur.com/AmItT.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AmItT.png" alt="skew"></a></p>
<p>They look like skew to me. What went wrong?</p>
| Hw Chu | 507,264 | <p>The idea is that, if you can show that the ratios of the rectangles are getting closer to <span class="math-container">$\sqrt 3$</span>, then none of them should be similar to each other. So you need to have some way to track <span class="math-container">$\frac{y_n}{x_n}$</span> in each iteration.</p>
<p>If you try some related expressions, you might discover that tracking <span class="math-container">$y_n^2 - 3x_n^2$</span> actually helps you a bit. Playing with the iteration relations you will get:</p>
<p><span class="math-container">$$
y_n^2 - 3x_n^2 = -2(y_{n-1}^2 - 3x_{n-1}^2).
$$</span></p>
<p>From this,
<span class="math-container">$$
\left|\left(\frac{y_n}{x_n}\right)^2-3\right| = 2\left(\frac{x_{n-1}}{x_n}\right)^2\left|\left(\frac{y_{n-1}}{x_{n-1}}\right)^2-3\right|.
$$</span></p>
<p>This leads to our assertion. Can you finish up the arguments?</p>
|
85,351 | <p>It has been proven that:</p>
<p>1) if $s$ is a non trivial zero $\rho$ of $\zeta(s)$ then so is $1−s$.</p>
<p>2) $\zeta(s) = 2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s) \zeta(1-s)$</p>
<p>3) $ 0 < \Re(\rho) <1$</p>
<p>From this it follows that when $s \to \rho$:</p>
<p>$\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}| = |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$</p>
<p>It is easy to see that the outcome will be $1$ for all $y$ in $s=\frac12 + y i$.</p>
<p>But if a $\rho$ would lie off this critical line, it also must reside in 'spots' where $\displaystyle \lim_{s \to \rho} |\dfrac{\zeta(s)}{\zeta(1-s)}|=1$.</p>
<p>On which points off the critical line could this occur? I found a surprisingly small domain (no proof).</p>
<p>The blue line shows the only values where:</p>
<p>$\displaystyle |2^s \pi^{s-1} \sin(\frac{\pi s}{2}) \Gamma(1-s)|=1$, $s=x + y i$, $ 0 \le x \le 1$.</p>
<p>Note that $y \to 2\pi$ for both $x=0$ and $x=1$. The $y$ rises only a little in the middle.</p>
<p>This doesn't say anything about whether or not off-line $\rho$'s are actually hiding on this curve. There still is an infinite number to check. However, I wondered if anything more is known about this curve? </p>
<p><img src="https://i.stack.imgur.com/95HcV.jpg" alt=""> <a href="https://web.archive.org/web/20131030092719if_/http://img822.imageshack.us/img822/3065/riemanntest.jpg" rel="nofollow noreferrer"><sup>(source: Wayback Machine)</sup></a></p>
| Community | -1 | <p>I do not think there is a nice solution. But the problem can be simplified a bit. First the condition that $f$ is bounded is irrelevant since you can always compose with $\arctan$. Second, the condition that $f$ is a function into $\mathbb{R}$ is not important, and in fact the only thing you need from $f$ is its kernel, i.e. the equivalence relation on $S\times S$:
$(a,b)\sim (c,d)$ iff $f(a,b)=f(c,d)$. Thus the problem is this: suppose that we are given an equivalence relation $\sim$ on $S\times S$ such that $(a,b)\sim (c,d)\to (b,a)\sim (d,c)$. When will there exists an associative commutative operation $\cdot$ such that $(a,b)\sim (c,d)$ if $a\cdot b=c\cdot d$. One can of course use the structure theory of commutative semigroups (these are more complicated than commutative groups but still manageable) to obtain
more information but I am not sure very much can be achieved. </p>
|
2,526,695 | <p>I've got following sequence formula:
$ a_{n}=2a_{n-1}-a_{n-2}+2^{n}+4$</p>
<p>where $ a_{0}=a_{1}=0$</p>
<p>I know what to do when I deal with sequence in form like this:</p>
<p>$ a_{n}=2a_{n-1}-a_{n-2}$
- when there's no other terms but previous terms of the sequence.
Can You tell me how to deal with this type of problems?
What's the general algorithm behind solving those?</p>
| robjohn | 13,854 | <p>Let $Sa_n=a_{n+1}$ be the shift operator on sequences. Then your equation becomes
$$
\left(1-S^{-1}\right)^2a_n=2^n+4\tag1
$$
where $1-S^{-1}$ is the <a href="https://en.wikipedia.org/wiki/Finite_difference#Forward.2C_backward.2C_and_central_differences" rel="nofollow noreferrer">backward difference operator</a>.</p>
<p>As with integration, we have a set of basic forms that can be validated by repeated backward difference:
$$
\left(1-S^{-1}\right)^2n^2=2\tag2
$$
$$
\left(1-S^{-1}\right)^22^n=2^{n-2}\tag3
$$
$$
\left(1-S^{-1}\right)^2a_n=0\implies a_n=C_1n+C_2\tag4
$$
looking at $(2)$, $(3)$, and $(4)$, we see that
$$
a_n=2^{n+2}+2n^2+C_1n+C_2\tag5
$$
Plugging in the conditions that $a_0=a_1=0$, we get
$$
a_n=2^{n+2}+2n^2-6n-4\tag6
$$</p>
|
13,882 | <p>Background: When Ueno builds the fully faithful functor from Var/k to Sch/k he mentions that the variety $V$ can be identified with the rational points of $t(V)$ over $k$. I know how to prove this on affine everything and will work out the general case at some future time.</p>
<p>The question that this got me thinking about was if $X$ is a $k$-scheme where $k$ is algebraically closed, then are the $k$-rational points of $X$ just the closed points? This is probably extremely well known, but I can't find it explicitly stated nor can I find a counterexample.</p>
<p>For $k$ not algebraically closed, I can come up with examples where this is not true. So in general is there some relation between the closed points and rational points on schemes (everything over $k$)?</p>
<p>This would give a bit more insight into what this functor does. It takes the variety and makes all the points into closed points of a scheme, then adds the generic points necessary to actually make it a legitimate scheme. General tangential thoughts on this are welcome as well.</p>
| Pete L. Clark | 1,149 | <p>The following result deals with the case of finite type affine schemes over an arbitrary field <span class="math-container">$k$</span>.</p>
<p>Theorem: Let <span class="math-container">$A$</span> be a finitely generated algebra over a field <span class="math-container">$k$</span>. Let <span class="math-container">$\iota: A \rightarrow \overline{A} = A \otimes_k \overline{k}$</span>.<br />
a) For every maximal ideal <span class="math-container">$\mathfrak{m}$</span> of <span class="math-container">$A$</span>, the set <span class="math-container">$\mathcal{M}(\mathfrak{m})$</span> of
maximal ideals <span class="math-container">$\mathcal{M}$</span> of <span class="math-container">$\overline{A}$</span> lying over <span class="math-container">$\mathfrak{m}$</span> is finite and
nonempty.<br />
b) The natural action of <span class="math-container">$G = \operatorname{Aut}(\overline{k}/k)$</span> on <span class="math-container">$\mathcal{M}(\mathfrak{m})$</span> is transitive. Thus <span class="math-container">$\operatorname{MaxSpec}(A) = G \backslash
\operatorname{MaxSpec}(\overline{A})$</span>.<br />
c) If <span class="math-container">$k$</span> is perfect, the size of the <span class="math-container">$G$</span>-orbit on <span class="math-container">$\mathfrak{m} \in \operatorname{MaxSpec}(A)$</span> is equal to the degree of the field extension of <span class="math-container">$k$</span> generated by
the coordinates in <span class="math-container">$\overline{k}^n$</span> of any <span class="math-container">$\mathcal{M}$</span> lying over <span class="math-container">$\mathfrak{m}$</span>.</p>
<p>In brief, the closed points correspond to the Galois orbits of the geometric points.</p>
<p>This is Theorem 8 in <a href="http://alpha.math.uga.edu/%7Epete/8320notes3.pdf" rel="nofollow noreferrer">http://alpha.math.uga.edu/~pete/8320notes3.pdf</a>.</p>
<p>The proof is left as an exercise, with some suggestions.</p>
<p>Exactly where this result came from, I cannot now remember. The text for the course that these notes accompany was Qing Liu's <em>Algebraic Geometry and Arithmetic Curves</em> (+1!), so it's a good shot that there is at least some cognate result in there.</p>
|
2,967,615 | <p>Given for example 2 functions,<span class="math-container">$\ n^{100} $</span> and<span class="math-container">$\ 2^n$</span>. I know that<span class="math-container">$\ 2^n$</span> grows faster and that therefore there is some<span class="math-container">$\ n$</span> where it will eventually overtake <span class="math-container">$\ n^{100} $</span> but how can I prove this, and also maybe find that<span class="math-container">$\ n$</span>?</p>
| DeepSea | 101,504 | <p>Solve <span class="math-container">$2^n > n^{100} \iff n\ln 2>100\ln n\iff \dfrac{n}{\ln n} > \dfrac{100}{\ln 2}$</span>. Let <span class="math-container">$n = 2^k$</span>, then <span class="math-container">$ \dfrac{n}{\ln n} = \dfrac{2^k}{k}$</span>,and you solve <span class="math-container">$2^k > 100k$</span>. Observe the first integer solution <span class="math-container">$k$</span> for this is <span class="math-container">$k = 10$</span>. Thus <span class="math-container">$n = 2^{10} = 1,024$</span> .</p>
|
2,967,615 | <p>Given for example 2 functions,<span class="math-container">$\ n^{100} $</span> and<span class="math-container">$\ 2^n$</span>. I know that<span class="math-container">$\ 2^n$</span> grows faster and that therefore there is some<span class="math-container">$\ n$</span> where it will eventually overtake <span class="math-container">$\ n^{100} $</span> but how can I prove this, and also maybe find that<span class="math-container">$\ n$</span>?</p>
| Mark Bennet | 2,906 | <p>One elementary way of proving that exponentials grow faster is to use the binomial expansion.</p>
<p>To illustrate, here <span class="math-container">$(1+1)^r=1+r+\binom r2+\binom r3+\dots$</span>, where we can choose <span class="math-container">$r$</span> large enough for all the terms we need to exist on the right-hand side.</p>
<p>To show that <span class="math-container">$2^r$</span> is eventually larger than <span class="math-container">$r^2$</span> we take the term <span class="math-container">$\binom r3=\frac {r(r-1)(r-2)}{3!}$</span> from the expansion and observe that this is a cubic in <span class="math-container">$r$</span> with positive leading term, and is therefore eventually greater than any quadratic in <span class="math-container">$r$</span> with positive leading term, and hence eventually greater that <span class="math-container">$r^2$</span>. Since this is just one of the positive terms on the right-hand side, we have for large enough <span class="math-container">$r$</span> <span class="math-container">$$2^r=(1+1)^r\gt\binom r3\gt r^2$$</span> with some very crude estimates. For <span class="math-container">$r^{100}$</span> we can use the same argument with <span class="math-container">$\binom r{101}$</span>, which is a polynomial of degree <span class="math-container">$101$</span> in <span class="math-container">$r$</span> and positive leading term, to give, for large enough <span class="math-container">$r$</span> <span class="math-container">$$2^r=(1+1)^r\gt\binom r{101}\gt r^{100}$$</span></p>
<p>The estimates here are very crude indeed, and are not good enough to give transition values for <span class="math-container">$r$</span>. But they are good enough to furnish a proof of the kind you wanted for the comparison of growth rates.</p>
|
1,918,071 | <p>Sometimes you will see theorems of the form "Let $H_1, \dots, H_n$. If $A$, then $B$". Sometimes "suppose" or "if" is used instead of "let". Here's an example:</p>
<ol>
<li><p>Let $x\in\mathbb{R}$. If $x\geq 0$, then $|x|=x$.</p></li>
<li><p>Suppose $x\in\mathbb{R}$. If $x\geq 0$, then $|x|=x$.</p></li>
<li><p>If $x\in\mathbb{R}$ and $x\geq 0$, then $|x|=x$.</p></li>
</ol>
<p>I'm under the impression that these are all equivalent ways of saying the same thing.
In this example, I would call "$x\in\mathbb{R}$" a hypothesis and "$x\geq 0$" the antecedent.
But in the third statement, is there an unambiguous contrapositive?
In certain contexts, I think it is understood that we're not really considering the case when $x\notin\mathbb{R}$. But "$x\in\mathbb{R}$" is nevertheless part of the antecedent in statement (3). So if we agree (1-3) are equivalent, then I see two contrapositives:</p>
<p>a) If $x\in\mathbb{R}$ and $|x|\neq x$, then $x<0$.</p>
<p>b) If $|x|\neq x$, then $x<0$ or $x\notin\mathbb{R}$. </p>
<p>I think Halmos' Naive Set Theory is an example where form (3) is preferred to (1,2). </p>
<p>The questions are:</p>
<ol>
<li><p>Are those statements equivalent?</p></li>
<li><p>In the third statement, what is The contrapositive? EDIT: Generally, if you see a theorem of the form "Let $H_1, \dots, H_n$. If $A$, then $B$", what is its contrapositive? How do you know?</p></li>
<li><p>Do mathematicians make any effort to separate the hypotheses ($H_1,\dots,H_n$) from the antecedent ($A$) of the claim? If so, how? Or is this one of those things everybody understands and no one is explicit about?</p></li>
</ol>
| Community | -1 | <p>$e^x (e^x+1) = 2$ does <strong>not</strong> imply that $e^x = 2$ or $e^x+1 = 2$. This type of reasoning only works if the right-hand side is zero. (This is why it's called the <a href="https://en.wikipedia.org/wiki/Zero-product_property" rel="nofollow">zero-product property</a>.)</p>
<p>In general, if $ab = 2$ then we could have $a=2$ and $b=1$, or $a=1$ and $b=2$, or $a=4$ and $b=1/2$, or $a=2\pi$ and $b=\frac1\pi$, etc. There are infinitely many possibilities if the right-hand side is not zero, which is why we need to use a different method in such cases.</p>
<p>The proper way to proceed with $e^{2x} + e^x - 2 = 0$ is to first rewrite $e^{2x}$ as $(e^x)^2$. Then the equation becomes
$$ (e^x)^2 + e^x - 2 = 0.$$
Now we can make the substitution $y = e^x$ to get
$$ y^2 + y - 2 = 0.$$</p>
<p>So now it's a plain quadratic equation. Factor to get $(y+2)(y-1) = 0$. Now we can use the zero-product property to say $y = -2$ or $y=1$. Back-substitute to get
$$ e^x = -2 \qquad \text{ or } \qquad e^x = 1.$$
Since $e^x$ can't be negative for any real value of $x$, the only valid case is $e^x = 1$, and this happens when $x = 0$.</p>
|
2,607,090 | <p>I have a function for which I know:</p>
<p>$f(2) = 2x -3y \\
f(3) = 5x - 6y \\
f(4) = 9x - 10 y \\
f(5) = 14x - 15y$</p>
<p>Assuming that $f$ is a polynomial, how do I find the general expression for $f$? After many minutes of fiddling I eventually found that this general expression works:</p>
<p>$f(N) = \frac{N(N+1)-2}{2}x - \frac{N(N+1)}{2}y$.</p>
<p>It's easy to verify that the expression works, but I found this by trial-and-error and I don't know if it's either unique or the simplest solution.</p>
| fleablood | 280,126 | <p>If you are assuming there is a pattern and that </p>
<p>$f(n) = A_n x - B_n y$ then all you are asking is:</p>
<p>If $A_2= 2; A_3 = 5; A_4 = 9; A_5 = 14$ what is $A_n$? And if $B_2= 3;B_3= 6; B_4 = 10; B_5 = 15$ what is $B_n$?</p>
<p>Well, there is no answer as just because something follows a pattern for $4$ numbers doesn't mean it follows then pattern forever.</p>
<p>But <em>if</em> we asuume the pattern we see that $A_k = A_{k-1} + k$ and $B_k = B_{k-1} + k$ continues forever we have:</p>
<p>So $A_n = A_{n-1} + n = A_{n-2} + (n-1) + n = ..... = A_0 + 1 + 2 +3+.....+n$.</p>
<p>$= A_0 + \sum\limits_{k=0}^n k = A_0 + \frac {n(n+1)}2$ [$*$]</p>
<p>And likewise $B_n = B_0 + \frac{n(n+1)}2$.</p>
<p>So if $A_2 = 2$ Then $A_2 = A_1 +2 = 2$ so $A_1 = 0$ and $A_1 = A_0 + 1 = 1$ so $A_0 = -1$. so $A_n = \frac {n(n+1)}2 - 1$.</p>
<p>And if $B_2 = 3$ than $B_2 =B_1 +2$ so $B_1 = 1$ and $B_0 = 0$.</p>
<p>So $B_n = \frac {n(n+1)}2$.</p>
<p>So $f(n) = x(\frac {n(n+1)}2) - y(\frac {n(n+1)}2)$</p>
<p><em>IF</em> we assume the pattern that we and $nx - ny$ to each step <em>always</em> holds true. (And we have <em>NO</em> reason on earth to assume that at all.)</p>
<hr>
<p>[$*$] It is a very well known result that $0 + 1 + 2+ 3 + ....... + n = \sum\limits_{k=0}^n k = \frac {n(n+1)}2$.</p>
<p>I'm not going to explain it hear but leave it as an exercise. (It is a rite of passage of all mathematicians to prove this result and every mathematician has stories and memories of where they were when they first learned that.) </p>
|
3,290,199 | <p>If I throw a fair dice <span class="math-container">$12$</span> times, the expected number of <span class="math-container">$6$</span> is <span class="math-container">$2$</span> i.e <span class="math-container">$6$</span> is expected to appear <span class="math-container">$2$</span> times when the dice is thrown <span class="math-container">$12$</span> times. But the probability of getting <span class="math-container">$6$</span> exactly <span class="math-container">$2$</span> times is <span class="math-container">${12}\choose{2}$$(1/6)^{2} (5/6)^{10} $</span> which is less than <span class="math-container">$1$</span>. </p>
<p>Now my question is <strong>How can you expect the face value six to appear for two times , when the possibility of that appearing for two times is very low?</strong></p>
<p>I am tying to give an analogy..If you are participating a game where you can win , lose or remain undecided. How can you expect to win When you know the possibility of winning the game is very low?</p>
<p>Can anyone please make me understand where I am getting wrong? I am really trying hard to understand.</p>
| José Carlos Santos | 446,262 | <p>The existence of an antiderivative and being integrable are distinct (although related) concepts.</p>
<p>Take<span class="math-container">$$\begin{array}{rccc}f\colon&[0,1]&\longrightarrow&\mathbb R\\&&x\mapsto&\begin{cases}x^2\sin\left(\frac1{x^2}\right)&\text{ if }x>0\\0&\text{ if }x=0.\end{cases}\end{array}$$</span>Then <span class="math-container">$f$</span> is differentiable, but <span class="math-container">$f'$</span> is unbounded. But, in particular, <span class="math-container">$f$</span> is an antiderivative of <span class="math-container">$f'$</span>.</p>
|
2,244,423 | <p>The function given is $f(x) = \sqrt[3]{{x}^2(2-x)}$.</p>
<p>Can anybody help me to find all asymptotes of this function. I know it doesn't have a vertical asymptote and I know that it's horizontal asymptote is $\sqrt[3]{-1}$, but I don't know how to find asymptote of the slope.</p>
<p>I'd prefer if someone could help me solving it using the formula given below:
$y = kx + l$ where $k = lim_{n\to\infty} \dfrac{f(x)}{x}$ and $l=lim_{n\to\infty}[f(x)-kx]$. I found $k$ that is $k=-1$ but I don't know how to find $l$.</p>
| Minz | 435,601 | <p>$\sqrt[3]{x^2(2-x)}=-x\sqrt[3]{1-\frac2x}=\boxed{\text{via Taylor}}=-x(1-\frac2{3x}+o(\frac1x))=-x+\frac23+o(1)$</p>
<p>As $\sqrt[3]{x^2(2-x)}- (-x+\frac23)$ tends to $0$ the asymptotes at $-\infty$ and $+\infty$ are $y=-x+\frac23$ according to definition.</p>
<p>Alternative approach is standart:
firstly search $k=\lim\limits_{x\to\pm\infty}\frac {f(x)}{x}$ and secondly define $b=\lim\limits_{x\to\pm\infty}(f(x)-kx)$. If both calculations would successful then the asymptotes would be $y=kx+b.$ They may be different at $-\infty$ and $+\infty$ or exist only at one of the infinities.</p>
|
4,037,295 | <p>The Cauchy Schwarz inequality says
<span class="math-container">$$
(ax+by+cz)^2 = (a^2+b^2+c^2)(x^2+y^2+z^2).
$$</span></p>
<p>I found that there is a kind of analogous inequality for <span class="math-container">$(ax+by+cz)^n$</span>
<span class="math-container">$$
(x+y+z)^n \leq 3^n(x^n+y^n +z^n).
$$</span>
if I remember it correctly.</p>
<p>How to prove this inequality?</p>
| WimC | 25,313 | <p>The factor can be improved to <span class="math-container">$3^{n-1}$</span>. For odd <span class="math-container">$n$</span> you have to assume probably that <span class="math-container">$x,y,z \geq 0$</span>. Then this is <a href="https://en.wikipedia.org/wiki/Jensen%27s_inequality" rel="nofollow noreferrer">Jensen’s inequality</a> for the monomial function <span class="math-container">$x \mapsto x^n$</span>: <span class="math-container">$$\left(\frac{x+y+z}3\right)^n \leq \frac{x^n + y^n + z^n}3.$$</span></p>
|
122,293 | <p>Let's consider all possible permutations of N numbers. Suppose for each permutation we calculate the sum of absolute differences between consecutive elements. Thus, for (1,2,3) one would have abs(1-2)+abs(2-3)=2. Is it possible to obtain a distribution of such sums for given N? For instance, for N=3 one would have 3!=6 permutations and possible sums as 2 and 3. The number of sums of 2 is 2 and for 3 it's equal 4. </p>
| Douglas Zare | 2,954 | <p>Andrew King pointed out that $E[X]$ is $(n^2-1)/3$. </p>
<p>So, to calculate the variance $E[X^2] - E[X]^2$ we need to find $E[X^2]$.</p>
<p>Let $\delta(i) = |\pi(i+1)-\pi(i)|$, so $X = \sum_{i=1}^{N-1} \delta(i)$.</p>
<p>$$\begin{eqnarray}X^2 &=& \bigg(\sum_{i=1}^{N-1} \delta(i)\bigg)^2 \newline &=& \sum_{i=1}^{N-1} \delta(i)^2 + 2 \sum_{i=1}^{N-2} \delta(i)\delta(i+1) + \sum_{|i-j|\gt 1} \delta(i)\delta(j) \end{eqnarray}$$</p>
<p>We can compute the expected values of each of these terms to find $E[X^2]$. By symmetry, we only need to find $(N-1)E[\delta(1)^2]$, $2(N-2)E[\delta(1)\delta(2)]$, and $((N-2)(N-3))E[\delta(1)\delta(3)].$</p>
<p>$$\begin{eqnarray}E[\delta(1)^2] &=& \frac{1}{N(N-1)} ~\sum_{a,b} (b-a)^2 \newline &=& \frac{1}{N(N-1)} \frac{N^4-N^2}{6} \newline &=&\frac{n^2+n}{6}\end{eqnarray}$$</p>
<p>I used Mathematica for the sums.</p>
<p>$$\begin{eqnarray} E[\delta(1)\delta(2)] &=& \frac{1}{N(N-1)(N-2)} \sum_{a,b,c~ \text{distinct}} |b-a||c-b| \newline &=& \frac{1}{N(N-1)(N-2)}\bigg(\sum_{a,b,c} |b-a||c-b| - \sum_{a,b,c | a=c} (b-a)^2 \bigg) \newline &=& \frac{1}{N(N-1)(N-2)} \bigg( \frac{7N^5 - 15N^3 - 8N}{60} - \frac{N^4-N^2}{6} \bigg)\newline &=& \frac{7N^2+11N+4}{60}\end{eqnarray}$$</p>
<p>The diagonal terms evaluate to $0$ or resemble the first sum.</p>
<p>$$\begin{eqnarray} E[\delta(1)\delta(3)] &=& \frac{1}{N(N-1)(N-2)(N-3)} \sum_{a,b,c,d ~ \text{distinct}} |b-a||d-c|\newline &=& \frac{1}{N(N-1)(N-2)(N-3)} \bigg( \sum_{a,b,c,d} |b-a||d-c| \newline & & - 4 \sum_{a,b,c,d | b=d} |b-a||c-b| + 2 \sum_{a,b,c,d|a=c,b=d} (b-a)^2 \bigg) \newline &=& \frac{1}{N(N-1)(N-2)(N-3)} \bigg( \frac{N^6-2N^4 + N^2}{9} \newline & & - 4\frac{7N^5-10N^4-15N^3+10N^2+8N}{60}+ 2 \frac{N^4-N^2}{6} \bigg) \newline &=& \frac{5N^5 - 21 N^4 + 35N^3+45N^2-40N -24}{45(N-1)(N-2)(N-3)} \newline &=& \frac{5N^4-16N^3+19N^2+64N+24}{45(N-2)(N-3)}\end{eqnarray}$$</p>
<p>If I calculated correctly, then </p>
<p>$$\begin{eqnarray}E[X^2] &= &\frac{N^3-N}{6} + \frac{7N^3-3N^2-18N-8}{30} \newline & & + \frac{5N^4-16N^3+19N^2+64N+24}{45} \newline &=& \frac{10N^4+4N^3+29N^2+59N+24}{90}\end{eqnarray}$$</p>
<p>$$\begin{eqnarray}\text{Var}[X] &=& E[X^2] -E[X]^2 \newline &=& \frac{4N^3+49N^2+59N+14}{90} \end{eqnarray}$$</p>
<p>For $N=100$, the standard deviation is $\sqrt\frac{249773}{5}=223.505$.</p>
<p>I believe a similar (but simpler) technique can be used to approximate higher moments of $X$. The dominant term is where no differences are adjacent, and the moments should be approximately the same as for random functions to $\lbrace 1, ..., n \rbrace$ instead of permutations. I think the method of moments would allow one to prove that the distribution is asymptotically normal.</p>
|
1,683,238 | <p>Let $[x]$ denote the fractional part of x. I'm quite lost about how to solve this problem. I suspect the solution is elementary, but all I can determine is that $x\notin\Bbb{Q}$.</p>
| John Wayland Bales | 246,513 | <p>Have you tried proof by contradiction? Suppose there is a positive $\alpha$ less than 1 such that for all positive $x$ less than 1 there exists an $n\in\mathbb{N}$ such that $\alpha^n\ge[nx]$. Perhaps for $x$ some function of $\alpha$ this leads to a contradiction? Such as perhaps $x=1-\alpha$ although that is just a random suggestion.</p>
|
237,838 | <p>The data are for the model $T(t) = T_{s} - (T_{s}-T_{0})e^{-\alpha t}$,
where $T_0$ is the temperature measured at time 0, and $T_{s}$ is the temperature at time $t=\infty$, or the environment temperature. $T_{s}$ and $\alpha$ are parameters to be determined.</p>
<p>How can I fit my data against this model? I'm trying to solve $T_{s}$ by $T_{s}=(T_{0}T_{2}-T_{1}^{2})/(T_{0}+T_{2}-2T_{1})$, where $T_{1}$ and $T_{2}$ are measurements in time $\Delta t$ and $2\Delta t$, respectively.</p>
<p>However, the results are varying a lot through the whole data set.</p>
<p>Shall I try gradient descent for the parameters?</p>
| Community | -1 | <p>Gradient descent might be overkill. </p>
<p>For convenience, use a temperature scale translated so that $T_0=0$ and the model is</p>
<p>$$T(t)=T_s(1-e^{-\alpha t}).$$</p>
<p>You want to minimize </p>
<p>$$E=\sum_i(T_i-T_s(1-e^{-\alpha t_i}))^2.$$</p>
<p>Setting an arbitrary value for $\alpha$, the least-squares estimate of $T_s$ is given by</p>
<p>$$\hat T_s(\alpha)=\frac{\sum_iT_i(1-e^{-\alpha t_i})}{\sum_i(1-e^{-\alpha t_i})^2},$$</p>
<p>from which you deduce</p>
<p>$$\hat E(\alpha)=\sum_i(T_i-\hat T_s(1-e^{-\alpha t_i}))^2.$$</p>
<p>The optimal $\alpha$ is found by unidimensional optimization.</p>
|
356,574 | <blockquote>
<p>$$\aleph_2^{\aleph_0}=\aleph_2$$</p>
</blockquote>
<p>Appreciate your help</p>
| Asaf Karagila | 622 | <p>One can use Hausdorff's formula (if it is for their disposal),</p>
<blockquote>
<p>$$\aleph_{\alpha+1}^{\aleph_\beta}=\aleph_\alpha^{\aleph_\beta}\cdot\aleph_{\alpha+1}$$</p>
</blockquote>
<p>From there we have: $$\aleph_2^{\aleph_0}=\aleph_1^{\aleph_0}\cdot\aleph_2=\left(2^{\aleph_0}\right)^{\aleph_0}\cdot\aleph_2=2^{\aleph_0}\cdot\aleph_2=\aleph_1\cdot\aleph_2=\aleph_2.$$</p>
<p>One may also derive Bernstein's formula: $\aleph_n^{\aleph_\beta}=2^{\aleph_\beta}\cdot\aleph_n$, from which the result is even more immediate.</p>
|
3,370,076 | <p>The total mechanical energy is conserved when a ball is dropped from a height of 4.00 <span class="math-container">$\mathit{m}$</span>, and it makes a elastic collision with the ground. Assuming no non-conservative forces are acting find the period of the ball. g of course is 9.81.</p>
<p><span class="math-container">\begin{align}
PE_g &= U_s \\
mgh &= \frac{1}2 kA^2 \\
mgh &= \frac{1}2 kh^2 \\
2mgh &= kh^2 \\
2\frac{g}{h} &= \frac{k}{m} \\
\omega &= \sqrt{\frac{k}{m}} = \sqrt{\frac{2g}{h}} \\
T &= \frac{2 \pi}{\omega}=2\pi \sqrt{\frac{h}{2g}} = \sqrt{2} \pi\sqrt{\frac{h} {g}}=2.837 s
\end{align}</span></p>
<p>Is my approach correct?</p>
<h2>Fixed Approach</h2>
<p><span class="math-container">\begin{align}
mgh &= \frac{1}2 m v^2_f \\
v_f &= \sqrt{2gh} \\
\frac{v_f - v_0}{g} &= t = \frac{T}{2} \\
2t &= T = 1.80 s
\end{align}</span></p>
| Danny Pak-Keung Chan | 374,270 | <p>It is trivial. Let <span class="math-container">$x_{1},x_{2}\in G$</span>. If <span class="math-container">$\alpha(x_{1})=\alpha(x_{2})$</span>,
then <span class="math-container">$ax_{1}=ax_{2}\Rightarrow a^{-1}(ax_{1})=a^{-1}(ax_{2})\Rightarrow x_{1}=x_{2}$</span>.
Therefore <span class="math-container">$\alpha$</span> is injective.</p>
<p>Let <span class="math-container">$y\in G$</span>. Define <span class="math-container">$x=a^{-1}y\in G$</span>,
then <span class="math-container">$\alpha(x)=a(a^{-1}y)=y$</span>, so <span class="math-container">$\alpha$</span> is surjective.</p>
|
301,264 | <p>Note: There is another question of the same title, but it is different and asks for group theory prerequisites in algebraic topology, while i want the topology prerequisites. </p>
<p>I am a physics undergrad, and I wish to take up a course on Introduction to Algebraic Topology for the next sem, which basically teaches the first two chapters of Hatcher, on Fundamental Group and Homology. However, I don't have a formal mathematics background in point-set topology, and I don't have enough time to go though whole books such as Munkres. So What part of point set topology from Munkres is actually used in the first two chapters of Hatcher?</p>
<p>More importantly, I wanted to know if the first chapter of the book <a href="http://rads.stackoverflow.com/amzn/click/1441972536">Topology, Geometry and Gauge Fields by Naber</a> or first 2 chapters of Lee's Topological Manifolds would be sufficient to provide me the necessary background for Hatcher.</p>
<p>Thanks in advance!</p>
| sponsoredwalk | 43,502 | <p>Chapter 1 of Hatcher corresponds to chapter 9 of Munkres. <a href="http://www.ictp.tv/diploma/search07-08.php?activityid=MTH&course=Topology">These</a> topology video lectures (syllabus <a href="http://diploma.ictp.it/courses/math/topology-top">here</a>) do chapters 2, 3 & 4 (topological space in terms of open sets, relating this to neighbourhoods, closed sets, limit points, interior, exterior, closure, boundary, denseness, base, subbase, constructions [subspace, product space, quotient space], continuity, connectedness, compactness, metric spaces, countability & separation) of Munkres before going on to do 9 straight away so you could take this as a guide to what you need to know from Munkres before doing Hatcher, however if you actually look at the subject you'll see chapter 4 of Munkres (questions of countability, separability, regularity & normality of spaces etc...) don't really appear in Hatcher apart from things on Hausdorff spaces which appear only as part of some exercises or in a few concepts tied up with manifolds (in other words, these concepts may be being implicitly assumed). Thus basing our judgement off of this we see that the first chapter of Naber is sufficient on these grounds... However you'd need the first 4 chapters of Lee's book to get this material in, & then skip to chapter 7 (with 5 & 6 of Lee relating to chapter 2 of Hatcher). </p>
<p>There's a crazy amount of abstract algebra involved in this subject (an introduction to which you'll find after lecture 25 in <a href="http://www.youtube.com/course?list=EC6763F57A61FE6FE8">here</a>) so I'd be equally worried about that if I didn't know much algebra.</p>
<p><a href="http://www.ictp.tv/diploma/search07-08.php?activityid=MTH&course=Algebraic_Topology&order=olderfirst">These video lectures</a> (syllabus <a href="http://diploma.ictp.it/courses/math/algebraic-topology-alt">here</a>) follow Hatcher & I found the very little I've seen useful mainly for the motivation the guy gives. If you download the files & use a program like <a href="http://www.irfanview.com/">IrfanView</a> to view the pictures as you watch the video on <a href="http://www.videolan.org/vlc/index.html">vlc player</a> or whatever it's much more bearable since you can freeze the position of the screen on the board as you scroll through 200 + pictures.</p>
<p>I wouldn't recommend you treat point set topology as something one could just rush through, I did & suffered very badly for it...</p>
|
3,406,106 | <p>Want to prove rigorously (if possible, since I was not able to think of any counter-example) that <span class="math-container">$\lim_{x\to a} f(x)$</span> exists <span class="math-container">$\implies \lim_{x\to a} f(x^2)$</span> exists. (I also have a feeling that the limits equate.)</p>
<p>I started with the <span class="math-container">$\epsilon-\delta$</span> definition of <span class="math-container">$\lim_{x\to a} f(x) = l$</span> which states that <span class="math-container">$\forall \epsilon >0 \exists \delta>0 \forall x:0<|x-a|<\delta \implies |f(x)-l| < \epsilon$</span>.</p>
<p>Now I want to show that <span class="math-container">$\forall \epsilon >0 \exists \delta>0 \forall x:0<|x-a|<\delta \implies |f(x^2)-l_1| < \epsilon$</span> where <span class="math-container">$l_1 \in \mathbb{R}$</span>.</p>
<p>I've got no clue how to continue. All help is appreciated.</p>
<p>As an additional note: I believe the converse (<span class="math-container">$\lim_{x\to a} f(x^2)$</span> exists <span class="math-container">$\implies \lim_{x\to a} f(x)$</span> exists) is not true as <span class="math-container">$f(x)=\frac{|x|}{x}$</span> is a counter-example.</p>
<p>Edit: A counter-example was found for the statement but I was wondering whether it is true for <span class="math-container">$a = 0$</span>? So in other words, I would like to prove or disprove the statement <span class="math-container">$\lim_{x\to 0} f(x)$</span> exists <span class="math-container">$\implies \lim_{x\to 0} f(x^2)$</span> exists</p>
| user | 505,767 | <p>It is always true for <span class="math-container">$a=0$</span> and <span class="math-container">$a=1$</span> otherwise just assume a discontinuous function at <span class="math-container">$x=a^2$</span>.</p>
|
3,522,736 | <p>I've been messing around with trying to negate this statement using DeMorgans laws and I keep ending up with incorrect answers such as (~p or ~q) and ~r. If someone could help me with the negation of compound statements. </p>
<p>Thank you.</p>
| Martin Argerami | 22,857 | <p>Since <span class="math-container">$x>0$</span>, you have <span class="math-container">$1+x>0$</span>, and <span class="math-container">$\frac1{1+x}>0$</span>. </p>
<p>And <span class="math-container">$1+x>1$</span>, so <span class="math-container">$\frac1{1+x}<1$</span>. </p>
<p>Equality is not achieved: if <span class="math-container">$\frac1{1+x}=1$</span>, then <span class="math-container">$x=0$</span>. And <span class="math-container">$\frac1{1+x}=0$</span> is impossible. </p>
<p>The existence of the limits, together with continuity, guarantees that you can find values as close to <span class="math-container">$0$</span> and <span class="math-container">$1$</span> as you want. So the range is <span class="math-container">$(0,1)$</span>. </p>
|
3,669,539 | <p>I’ve been looking at non-trivial solutions of ODEs. I found one and have problems with it.</p>
<p><span class="math-container">$y’(x)=\frac{1}{4}\sqrt{y(x)}$</span></p>
<p><span class="math-container">$y(0)=0$</span></p>
<p>I know the one of the solutions of this ODE is</p>
<p><span class="math-container">$y(x) = \frac{x^2}{64}$</span></p>
<p>Are there any other non-trivial solutions to this ODE. Thanks</p>
| Bernard | 202,857 | <p>This is an antiderivative problem. The fundamental theorem of integral calculus asserts that
<span class="math-container">$$y(x)=\int_0^x\tfrac14\sqrt{ t\mkern1mu}\mkern3mu\mathrm dt=\tfrac14\tfrac23t^{3/2}\bigg|_0^x{}=\tfrac16 x^{3/2}.$$</span></p>
|
497,422 | <p>Which is bigger: $a$ or $a^2$ and what is the proof of that?</p>
<p>I'm kinda stuck and because there are cases where $a$ is bigger and other cases where $a^2$ is bigger.</p>
| mrf | 19,440 | <p>Hint: Factorize $a^2 - a$. When is this number greater than or smaller than $0$?</p>
|
1,106,325 | <p>In <a href="https://www.encyclopediaofmath.org/index.php/Codimension" rel="nofollow noreferrer">this article about codimension</a> there is the following remark: </p>
<p>The codimension of a subspace $L$ of a vector space $V$ is equal to the dimension of any complement of $L$ in $V$, since all complements have the same dimension (as the orthogonal complement).</p>
<p>I know the definition of orthogonal complement in a Hilbert space but it's not clear to me how the orthogonal complement is defined in an arbitrary vector space. </p>
<blockquote>
<p>How to define it in this case?</p>
</blockquote>
<p>Also, </p>
<blockquote>
<p>What's the definition of a (non-orthogonal) complement of a subspace?</p>
</blockquote>
<p><strong>Edit</strong></p>
<p>"complement" is also used in <a href="https://math.stackexchange.com/a/2678/167889">this answer here</a>. It's really not clear to me how to define a complement (if it's not orthogonal).</p>
| user1801328 | 147,676 | <p>The most general statement of orthogonality I've seen is in a Banach space. If you have a Banach space $X$ and denote it's dual by $X^{*} $ then, for $V \subseteq X$, the complement of $V $ is defined as all $x^{*} \in X^{*}$ s.t for every $v \in V $, $x^{*}(v)=0$. </p>
|
881,520 | <p>Take half a square with side length $1$.
The resulting right-angled triangle ABC
has two angles of $45^\circ$.
By Pythagoras’ theorem, the hypotenuse
AC has length $\sqrt{2}$. Applying the definitions on the previous page gives the values in the table below. that $\sin 30^\circ= \frac{1}{2}$</p>
<p>Sorry I cannot provide diagram, but from my understanding $\sin =$ opposite / hypotenuse. How is the value $0.5$ derived then? No possible combination. What point of reference should I be looking from?</p>
| Henry | 6,460 | <p>Take an equilateral triangle of side $1$. </p>
<p>Bisect it through a vertex and the midpoint of the opposite side. </p>
<p>You now have two right-angled triangles with angles $30^\circ, 60^\circ, 90^\circ$ with the edge opposite the $30^\circ$ of length $\frac12$ and hypotenuse $1$. So $$\sin (30^\circ)=\frac{\frac12}{1}=\frac12.$$</p>
|
550,659 | <blockquote>
<p>A space <span class="math-container">$X$</span> is locally metrizable if each point <span class="math-container">$x$</span> of <span class="math-container">$X$</span> has a neighborhood that is metrizable in the subspace topology. Show that a compact Hausdorff space <span class="math-container">$X$</span> is metrizable if it is locally metrizable.</p>
<p><strong>Hint:</strong> Show that <span class="math-container">$X$</span> is a finite union of open subspaces, each of which has a countable basis.</p>
</blockquote>
<p>I tried to use the fact of compact space. But I do not know if the opens are compact subspaces.</p>
| Berci | 41,488 | <p>Of course, if $X$ is metrizable then it is also locally metrizable. </p>
<p>For the other direction, you could really follow the hint. Supposed that $X$ is locally metrizable, there exists an open cover $\bigcup_{x\in X}U_x$ of $X$ where $U_x$ is a metrizable neighborhood of $x$. Because of compactness, it has a finite subcover: $\bigcup_{i<n} U_i=X$ where $U_i:=U_{x_i}$ for some $x_i$.</p>
<p>Because of metrizability, each $U_i$ has countable base $(V_{i,j})_j$ of open subsets, so that all finite intersections of these are still countable, and they give a base of $X$.</p>
<p>The space also has to satisfy the separation axiom $T_3$, but this holds as $X$ is Hausdorff and compact.</p>
|
550,659 | <blockquote>
<p>A space <span class="math-container">$X$</span> is locally metrizable if each point <span class="math-container">$x$</span> of <span class="math-container">$X$</span> has a neighborhood that is metrizable in the subspace topology. Show that a compact Hausdorff space <span class="math-container">$X$</span> is metrizable if it is locally metrizable.</p>
<p><strong>Hint:</strong> Show that <span class="math-container">$X$</span> is a finite union of open subspaces, each of which has a countable basis.</p>
</blockquote>
<p>I tried to use the fact of compact space. But I do not know if the opens are compact subspaces.</p>
| user558252 | 558,252 | <p>[Hint: Using normality/regularity]</p>
<p>The space $X$ is compact Hausdorff, so it is normal/regular.</p>
<p>Now for each $x\in X$, there is a metrizable open neighborhood $U_{x}$ of $x$, for this $U_{x}$, by regularity, there is an open neighborhood $V_{x}$ of $x$ such that
$x\in V_x \subset V_{x}^{-} \subset U_{x}$. Notice that the closure $V_x^{-}$ of $V_x$ is a closed subset of a compact space and a subspace of a metrizable space, therefore it is a compact metrizable space and has a countable basis, so $V_x$ has a countable basis. </p>
<p>Since $X$ is compact, and the family $\{V_x\}_{x\in X}$ of opens covers $X$, there is a finite subcover in which every member is open and has a countable basis. Hence $X$ has a countable basis.</p>
|
2,564,217 | <p>For a project I'm doing, I'm wrapping an led strip light around a tube. The tube is 19mm in diameter and 915mm tall. I'm going to coil the led strip around the tube from top to bottom and the strip is 8mm wide, so the coils will be 8mm apart. How long does the led strip need to be to fully cover the tube?</p>
<p>This reminds me of a popular question on Math SE about a toilet paper roll, but slightly different. I estimated this by measuring how many 8mm wide circles could fit around the tube, then multiplied by the circumference. However, I don't know how to calculate the exact length of the coil. Out of curiosity, how would you find the exact length of the coil wrapping around the tube with each coil being 8mm apart?</p>
| Graham Kemp | 135,106 | <blockquote>
<p>I drew a 6x6 grid and tried labeling all the possible outcomes followed by the number of times they occur, but I feel like I got no where. I know approach is very wrong.</p>
</blockquote>
<p>The approach is correct. The table should look as follows, just complete it.</p>
<p>$$\begin{array}{:c|c:c:c:c:c:c:}\hline \lower{1ex}X\backslash Y & 0 & 1 & 2 & 3 & 4 & 5\\\hline 1& 1/36& 1/36& 1/36& 1/36& 1/36& 1/36 \\ \hdashline 2 & 1/36 & 2/36 & 1/36& 1/36& 1/36 & 0 \\ \hdashline 3 & \\ \hdashline 4\\ \hdashline 5\\ \hdashline 6 \\ \hline\end{array}$$</p>
<p>Or perhaps start off by listing outcomes of the die rolls that lead to each pair of results.</p>
<p>$$\begin{array}{:c|c:c:c:c:c:c:}\hline \lower{1ex}X\backslash Y & 0 & 1 & 2 & 3 & 4 & 5\\\hline 1& (1,1)& (1,2)& (1,3)& (1,4)& (1,5)& (1,6) \\ \hdashline 2 & (2,2) & {(2,1),\\(2,3)} & (2,4)& (2,5)& (2,6) & \{\} \\ \hdashline 3 & \\ \hdashline 4\\ \hdashline 5\\ \hdashline 6 \\ \hline\end{array}$$</p>
|
2,956,791 | <p>There is an equation here:
<span class="math-container">$$\sqrt{x+1}-x^2+1=0$$</span>
Now we want to write the equation <span class="math-container">$f(x)$</span> like <span class="math-container">$h(x)=g(x)$</span> in a way that we know how to draw h and g functions diagram.
Then we draw the h and g function diagrams and find the common points of them. So it will be number of the <span class="math-container">$f(x)$</span> roots that here is the equation mentioned top.
Actually now my problem is with drawing the first equation's diagram
I want you to draw its diagrams like <span class="math-container">$\sqrt{x-1}$</span> syep by step. Please help me with it!</p>
| Dr. Sonnhard Graubner | 175,066 | <p>Hint: Write your equation in the form
<span class="math-container">$$\sqrt{x+1}=(x+1)(x-1)$$</span></p>
|
2,142,042 | <p>how would you use induction to prove this:</p>
<p>$\sin(x)-sin(3x)+sin(5x)-...+(-1)^{(n+1)}sin[(2n-1)x] = \frac{(-1)^{(n+1)}sin2nx}{2cosx} $</p>
<p>I know how you assume its true for n=k, and then prove for n=k+1, but I get to </p>
<p>Left Hand Side: $\frac{(-1)^{(k+1)}sin2kx}{2cosx}+(-1)^{k+2}sin[(2k+1)x]$ but I'm not sure what step to take next.</p>
<p>any help would be appreciated.
Cheers</p>
| Community | -1 | <p>First pull the constant $a$ and $b$ away by a scaling of the variable, and get rid of the square with $ax^2/b\to t$.</p>
<p>To a constant factor, you now have</p>
<p>$$\int_0^\infty\frac{t^{p-1}}{(t+1)^{p+q}}dt.$$</p>
<p>Now you convert to the classical Beta form by $u:=t/(t+1)$,</p>
<p>$$\int_0^1u^{p-1}(1-u)^{q-1}du.$$</p>
<hr>
<p>The value of this integral can be obtained using the Gamma function. The standard proof relies on the decomposition of a particular double integral as a product of integrals. It bears some resemblance to the convolution theorem of the Laplace transform.</p>
|
1,050,917 | <p>I have a problem that I have to solve. I need to find center of the circle containing the point $(x,y)$. The point is $x=2,y=3$ with radius $r=3$. I need to find the center of circle. Is there equation for that? I use this equation.<br>
$$(x-h)^2+(y-k)^2=r^2$$
How I can find $h$ and $k$ for the center of circle if I know the point on circle and the diameter of circle? </p>
| Archis Welankar | 275,884 | <p>Acoording to you $(x,y)$ is the centre with radius $3$ . So by the distance formula $(x-2)^2+(y-3)^2=3^2=9$. Then by plugging in any value for y you get a corresponding value for x. There are 2 unknowns and only 1 equation. So we need another parameter to solve for a specific point. Two, of the infinitely many, answers are for example: $(x,y)=(2,6)$ or $(2\sqrt{2},4)$.</p>
|
135,936 | <p>I need this one result to do a problem correctly.</p>
<p>I want to show that for any $b \in \mathbb{C}$ and $z$ a complex variable:</p>
<p>$$ |z^2 + b^2| \geq |z|^{2} - |b|^{2}$$ </p>
<p>My attempts have only led me to conclude that </p>
<p>$$ |z^2 + b^2| > \frac{|z|^{2} + |b|^{2}}{2}$$ </p>
| martini | 15,379 | <p>$$ |z|^2 = |z^2| = |z^2 + b^2 - b^2| \le |z^2 + b^2| + |b|^2 $$
and your inequality follows. I suppose you are missing an $|\cdot|$ arround the $b$? For you cannot compare complex numbers well.</p>
|
4,532,279 | <p>Let <span class="math-container">$k$</span> be an algebraically closed field and <span class="math-container">$X$</span> be an affine surface over <span class="math-container">$k$</span>. Suppose <span class="math-container">$\phi: \mathbb{A}^1 \to X$</span> is a non-constant morphism. Then we know that <span class="math-container">$Im(\phi)$</span> is closed in <span class="math-container">$X$</span>. Is it true that <span class="math-container">$\phi$</span> self intersects at finitely many points i.e. there are finitely many unordered pairs <span class="math-container">$(t_0, t_1)$</span> such that <span class="math-container">$\phi(t_0) = \phi(t_1)$</span>? I think it is true since there are finitely many points in <span class="math-container">$Im(\phi)$</span> where <span class="math-container">$\phi$</span> is not smooth. The singularity of <span class="math-container">$\phi$</span> is at those points precisely where <span class="math-container">$\phi$</span> intersects itself. <br />
Please give some hint. Thanks in advance.</p>
| Qiaochu Yuan | 232 | <p>Conrad's suggestion in the comments works great and is probably the most low-tech solution. Here are some others. Write <span class="math-container">$z = e^{i \theta}$</span>, so the question is to show that <span class="math-container">$z^n + z^{-n}$</span> is a polynomial in <span class="math-container">$z + z^{-1}$</span>. These are <a href="https://en.wikipedia.org/wiki/Laurent_polynomial" rel="nofollow noreferrer">Laurent polynomials</a> (polynomials in <span class="math-container">$z$</span> and <span class="math-container">$z^{-1}$</span>).</p>
<ol>
<li><p>Consider the vector space of Laurent polynomials <span class="math-container">$f(z)$</span> containing terms between <span class="math-container">$z^{-n}$</span> and <span class="math-container">$z^n$</span> and satisfying <span class="math-container">$f(z) = f(z^{-1})$</span>. It's not hard to see that <span class="math-container">$\{ z^k + z^{-k} : 0 \le k \le n \}$</span> is a basis for this vector space, so it has dimension <span class="math-container">$n+1$</span>. On the other hand, <span class="math-container">$\{ (z + z^{-1})^k : 0 \le k \le n \}$</span> is also a basis (e.g. because it consists of polynomials of different degrees so is linearly independent, and has size <span class="math-container">$n + 1$</span>). So there must be a change-of-basis matrix relating the two.</p>
</li>
<li><p>We have <span class="math-container">$(z + z^{-1})^n = \sum_{k \ge 0} {n \choose k} z^{n-2k}$</span>, which shows that <span class="math-container">$(\cos \theta)^n$</span> can be written as a linear combination of <span class="math-container">$\cos k \theta, 0 \le k \le n$</span> such that the coefficient of <span class="math-container">$\cos n \theta$</span> is <span class="math-container">$1$</span>. This gives a triangular matrix relating the sequences of functions <span class="math-container">$\cos^n \theta$</span> and <span class="math-container">$\cos n\theta$</span> with <span class="math-container">$1$</span>s on the diagonal, and all such matrices have an inverse which is another such matrix. It follows that <span class="math-container">$\cos n \theta$</span> is a monic polynomial in <span class="math-container">$\cos \theta$</span> of degree <span class="math-container">$n$</span>. This is basically a somewhat more explicit variant of the first argument.</p>
</li>
<li><p>Consider the generating function <span class="math-container">$\sum_{n \ge 0} (z^n + z^{-n}) t^n = \frac{1}{1 - zt} + \frac{1}{1 - z^{-1} t}$</span>. Adding these fractions gives <span class="math-container">$\frac{2 - (z + z^{-1}) t}{1 - (z + z^{-1}) t + t^2}$</span> and expanding this out produces a generating function in <span class="math-container">$t$</span> whose coefficients are polynomials in <span class="math-container">$z + z^{-1}$</span>.</p>
</li>
</ol>
<p>That last expression is, up to a factor of <span class="math-container">$2$</span>, the generating function of the <a href="https://en.wikipedia.org/wiki/Chebyshev_polynomials" rel="nofollow noreferrer">Chebyshev polynomials of the first kind</a> <span class="math-container">$T_n$</span>, the polynomials satisfying <span class="math-container">$\cos n \theta = T_n(\cos \theta)$</span> we've been trying to find, as mentioned by Jean Marie in the comments. Much is known about them and they have lots of interesting properties.</p>
<p>Overall I'd summarize the thrust of all the above arguments as saying that the vector space of Laurent polynomials satisfying <span class="math-container">$f(z) = f(z^{-1})$</span> (one might call them <em>symmetric</em>) has a basis given by any sequence <span class="math-container">$f_n(z)$</span> of symmetric Laurent polynomials of degree <span class="math-container">$n$</span>.</p>
<p><strong>Edit:</strong> Ah, I didn't see that the title asks for integer coefficients (the body only asks for complex coefficients). The first argument, as written, only provides rational coefficients, but the other two provide integer coefficients; the inverse of a triangular matrix with <span class="math-container">$1$</span>s on the diagonal involves no divisions, so for an integer such matrix the inverse also consists of integers. The generating function expansion also involves no divisions. The general statement above about Laurent polynomials is still true over <span class="math-container">$\mathbb{Z}$</span> if one adds the adjective "monic."</p>
|
4,532,279 | <p>Let <span class="math-container">$k$</span> be an algebraically closed field and <span class="math-container">$X$</span> be an affine surface over <span class="math-container">$k$</span>. Suppose <span class="math-container">$\phi: \mathbb{A}^1 \to X$</span> is a non-constant morphism. Then we know that <span class="math-container">$Im(\phi)$</span> is closed in <span class="math-container">$X$</span>. Is it true that <span class="math-container">$\phi$</span> self intersects at finitely many points i.e. there are finitely many unordered pairs <span class="math-container">$(t_0, t_1)$</span> such that <span class="math-container">$\phi(t_0) = \phi(t_1)$</span>? I think it is true since there are finitely many points in <span class="math-container">$Im(\phi)$</span> where <span class="math-container">$\phi$</span> is not smooth. The singularity of <span class="math-container">$\phi$</span> is at those points precisely where <span class="math-container">$\phi$</span> intersects itself. <br />
Please give some hint. Thanks in advance.</p>
| GEdgar | 442 | <p>Prove more: For all positive integers <span class="math-container">$n$</span>,
<span class="math-container">$$
\cos(n\theta) = T_n(\cos(\theta)),
\qquad \sin(n\theta) = \sin(\theta) U_{n-1}(\cos\theta)
$$</span>
where <span class="math-container">$T_n(x)$</span> and <span class="math-container">$U_{n-1}(x)$</span> are polynomials with integer coefficients.</p>
<p>Prove it by induction.<br />
<span class="math-container">$$
\cos(1\theta) = \cos(\theta) = T_1(\cos(\theta)),\quad
\sin(1\theta) = \sin(\theta) = \sin(\theta)U_0(\cos(\theta)),
$$</span>
where <span class="math-container">$T_1(x) = x$</span> and <span class="math-container">$U_0(x) = 1$</span> are polynomials with integer coefficients.</p>
<p>Let <span class="math-container">$n \ge 1$</span> and assume <span class="math-container">$\cos(n\theta) = T_n(\cos(\theta))$</span>
and <span class="math-container">$\sin(n\theta) = \sin(\theta)U_{n-1}(\cos(\theta))$</span>,
where <span class="math-container">$T_n(x)$</span> and <span class="math-container">$U_{n-1}(x)$</span> are a polynomials with integer coefficients. Then for <span class="math-container">$n+1$</span> we have
<span class="math-container">\begin{align}
\cos((n+1)\theta) &= \cos(n\theta)\cos(\theta) - \sin(n\theta)\sin(\theta)
\\ &= T_n(\cos(\theta)\cos(\theta) - \sin^2(\theta)U_{n-1}(\cos(\theta))
\\ &=T_{n+1}(\cos(\theta)),
\end{align}</span>
where <span class="math-container">$T_{n+1}(x) = xT_n(x)-(1-x^2)U_{n-1}(x)$</span> is a polynomial with integer coefficients; and
<span class="math-container">\begin{align}
\sin((n+1)\theta) &= \sin(n\theta)\cos(\theta)+\cos(n\theta)\sin(\theta)
\\ &= \sin(\theta)U_{n-1}(\cos(\theta))\cos(\theta)
+T_n(\cos(\theta))\sin(\theta)
\\ &=\sin(\theta) U_n(\cos(\theta)),
\end{align}</span>
where <span class="math-container">$U_n(x) = U_{n-1}(x)x + T_n(x)$</span> is a polynomial with integer coefficients.</p>
<hr />
<p>If you need a formula for the coefficients ... see references on "Chebyshev polynomials".</p>
|
14,385 | <p>I have always taught my students that the <span class="math-container">$y$</span>-intercept of a line is the <span class="math-container">$y$</span>-coordinate of the point of intersection of a line with the <span class="math-container">$y$</span>-axis, that is, for the line given by the equation <span class="math-container">$y=mx+y_0$</span>, the <span class="math-container">$y$</span>-intercept is <span class="math-container">$y_0$</span>. I emphasize that that the <span class="math-container">$y$</span>-intercept is the <em>number</em> <span class="math-container">$y_0$</span> and not the <em>point</em> <span class="math-container">$(0,y_0)$</span>.</p>
<p>But I was quite surprised when I recently looked at the <a href="https://en.wikipedia.org/wiki/Intercept" rel="nofollow noreferrer">Wikipedia</a> and <a href="http://mathworld.wolfram.com/x-Intercept.html" rel="nofollow noreferrer">Wolfram</a> <a href="http://mathworld.wolfram.com/y-Intercept.html" rel="nofollow noreferrer">MathWorld</a> entries for <span class="math-container">$y$</span>-intercept because these define the intercept as a point and not as a number ("the point where a line crosses the y-axis" and "The point at which a curve or function crosses the y-axis").</p>
<p>Further investigation yielded inconsistencies: the Wikipedia entry for "<a href="https://en.wikipedia.org/wiki/Line_(geometry)#On_the_Cartesian_plane" rel="nofollow noreferrer">Line (geometry)</a>" states that in the equation <span class="math-container">$y=mx+b$</span>, "<span class="math-container">$b$</span> is the y-intercept of the line"; the Wolfram MathWorld entry for "<a href="http://mathworld.wolfram.com/Line.html" rel="nofollow noreferrer">Line</a>" states that "The line with <span class="math-container">$y$</span>-intercept <span class="math-container">$b$</span> and slope <span class="math-container">$m$</span> is given by the slope-intercept form <span class="math-container">$y=mx+b$</span>.</p>
<hr />
<p><sup>Edit made on February 21, 2021</sup></p>
<p>According to the <em>Dictionary of Analysis, Calculus, and Differential Equations</em> (edited by Douglas N. Clark, published by CRC Press in 2000),</p>
<blockquote>
<p><strong>intercept</strong> The point(s) where a curve or graph of a function in <span class="math-container">$\mathbf R^n$</span> crosses one of the
axes. For the graph of <span class="math-container">$y=f(x)$</span> in <span class="math-container">$\mathbf R^2$</span>, the <span class="math-container">$y$</span>-<em>intercept</em> is the point <span class="math-container">$(0,f(0))$</span> and the <span class="math-container">$x$</span>-<em>intercepts</em> are the points <span class="math-container">$(p,f(p))$</span> such that <span class="math-container">$f(p)=0$</span>.</p>
</blockquote>
<p>Unfortunately, the book does not consistently use that definition.</p>
<blockquote>
<p><strong>slope-intercept equation of line</strong> An equation of the form <span class="math-container">$y=mx+b$</span>, for a straight line in <span class="math-container">$\mathbf R^2$</span>. Here <span class="math-container">$m$</span> is the slope of the line and <span class="math-container">$b$</span> is the <span class="math-container">$y$</span>-intercept; that is, <span class="math-container">$y=b$</span>, when <span class="math-container">$x=0$</span>.</p>
</blockquote>
<p>Thus, even though the book defines an intercept as a point, it uses the term to denote a number.</p>
<hr />
<p>Is there a trusted source targeted at mathematics educators (from, say, a government agency, an educational institution, or an organization) that defines "intercept" and consistently uses that definition?</p>
| Xander Henderson | 8,571 | <p>Some time spent with the Google does not seem to turn up very much. The best answer that I can find seems to be from the Common Core Standards, and is ambiguous, at best. On page 69 of the <a href="https://www.engageny.org/resource/new-york-state-p-12-common-core-learning-standards-for-mathematics" rel="noreferrer">Core Learning Standards</a> for mathematics (as provided by the State of New York; the same line appears in <a href="https://www.cde.ca.gov/be/st/ss/documents/ccssmathstandardaug2013.pdf" rel="noreferrer">California's document</a> on page 94), it states that students should be able to</p>
<blockquote>
<p>[i]nterpret the slope (rate of change) and the intercept (constant term) of a linear model in the context of the data.</p>
</blockquote>
<p>There are other references to intercepts, but they are ambiguous, at best. This seems to indicate that the Common Core regards intercepts as values, not points.</p>
<hr>
<p>A quick perusal of the books on my (and my office-mate's) shelves gives an equally ambiguous answer. Most (though not all) of the precalculus texts say that the line given by $y = mx + b$ has the value $b$ as it's $y$-intercept, while most of my calculus texts define the $y$-intercept to be the point $(0,b)$. A notable exception (notable as it is quite widely used) is a copy of the 6th edition of Stewart's <em>Calculus</em>, which gives the $y$-intercept as a value, not a point.</p>
|
14,385 | <p>I have always taught my students that the <span class="math-container">$y$</span>-intercept of a line is the <span class="math-container">$y$</span>-coordinate of the point of intersection of a line with the <span class="math-container">$y$</span>-axis, that is, for the line given by the equation <span class="math-container">$y=mx+y_0$</span>, the <span class="math-container">$y$</span>-intercept is <span class="math-container">$y_0$</span>. I emphasize that that the <span class="math-container">$y$</span>-intercept is the <em>number</em> <span class="math-container">$y_0$</span> and not the <em>point</em> <span class="math-container">$(0,y_0)$</span>.</p>
<p>But I was quite surprised when I recently looked at the <a href="https://en.wikipedia.org/wiki/Intercept" rel="nofollow noreferrer">Wikipedia</a> and <a href="http://mathworld.wolfram.com/x-Intercept.html" rel="nofollow noreferrer">Wolfram</a> <a href="http://mathworld.wolfram.com/y-Intercept.html" rel="nofollow noreferrer">MathWorld</a> entries for <span class="math-container">$y$</span>-intercept because these define the intercept as a point and not as a number ("the point where a line crosses the y-axis" and "The point at which a curve or function crosses the y-axis").</p>
<p>Further investigation yielded inconsistencies: the Wikipedia entry for "<a href="https://en.wikipedia.org/wiki/Line_(geometry)#On_the_Cartesian_plane" rel="nofollow noreferrer">Line (geometry)</a>" states that in the equation <span class="math-container">$y=mx+b$</span>, "<span class="math-container">$b$</span> is the y-intercept of the line"; the Wolfram MathWorld entry for "<a href="http://mathworld.wolfram.com/Line.html" rel="nofollow noreferrer">Line</a>" states that "The line with <span class="math-container">$y$</span>-intercept <span class="math-container">$b$</span> and slope <span class="math-container">$m$</span> is given by the slope-intercept form <span class="math-container">$y=mx+b$</span>.</p>
<hr />
<p><sup>Edit made on February 21, 2021</sup></p>
<p>According to the <em>Dictionary of Analysis, Calculus, and Differential Equations</em> (edited by Douglas N. Clark, published by CRC Press in 2000),</p>
<blockquote>
<p><strong>intercept</strong> The point(s) where a curve or graph of a function in <span class="math-container">$\mathbf R^n$</span> crosses one of the
axes. For the graph of <span class="math-container">$y=f(x)$</span> in <span class="math-container">$\mathbf R^2$</span>, the <span class="math-container">$y$</span>-<em>intercept</em> is the point <span class="math-container">$(0,f(0))$</span> and the <span class="math-container">$x$</span>-<em>intercepts</em> are the points <span class="math-container">$(p,f(p))$</span> such that <span class="math-container">$f(p)=0$</span>.</p>
</blockquote>
<p>Unfortunately, the book does not consistently use that definition.</p>
<blockquote>
<p><strong>slope-intercept equation of line</strong> An equation of the form <span class="math-container">$y=mx+b$</span>, for a straight line in <span class="math-container">$\mathbf R^2$</span>. Here <span class="math-container">$m$</span> is the slope of the line and <span class="math-container">$b$</span> is the <span class="math-container">$y$</span>-intercept; that is, <span class="math-container">$y=b$</span>, when <span class="math-container">$x=0$</span>.</p>
</blockquote>
<p>Thus, even though the book defines an intercept as a point, it uses the term to denote a number.</p>
<hr />
<p>Is there a trusted source targeted at mathematics educators (from, say, a government agency, an educational institution, or an organization) that defines "intercept" and consistently uses that definition?</p>
| Edwin F. Sampang | 14,866 | <p>I think y-intercept is a directed line segment. It is a vector in which it has a magnitude and direction. For example for the line y=2x-8, the y-intercept must be -8 meaning it is 8 units downward from the origin. The point involved is (0,-8) but it is not the y-intercept. It is only the point where the line crosses the y-axis. If you consider y-intercept as a point, what will happen to the equation of a line mentioned above. Will it be y=2x+(0,-8)?</p>
|
3,969,598 | <p>Let <span class="math-container">$f: \mathbb{R} \rightarrow \mathbb{R}$</span> a monotonically increasing function and <span class="math-container">$A \subset \mathbb{R}$</span> where <span class="math-container">$A \neq \emptyset$</span> and boundend.</p>
<p>i) If f is continuous function, prove that <span class="math-container">$f(\sup (A))= \sup (f(A))$</span></p>
<p>ii) Find a function <span class="math-container">$f: \mathbb{R} \rightarrow \mathbb{R}$</span> wich do not fulfilled i).</p>
<p>For i), I thought because <span class="math-container">$A$</span> is boundend it has supreme,<br />
<span class="math-container">$$ x \le \sup A$$</span>
In other hand, because <span class="math-container">$f$</span> is monotonically increasing, if <span class="math-container">$x \le \sup A$</span> then <span class="math-container">$f(x) \le f(\sup A)$</span> for <span class="math-container">$x \in A$</span>.</p>
<p>As <span class="math-container">$f$</span> is continuous and monotonically increasing in a boundend set, the set <span class="math-container">$f(A)$</span> has a supreme, i.e <span class="math-container">$f(y) \le \sup f(A)$</span> for <span class="math-container">$y \in A$</span>, so <span class="math-container">$\sup f(A)$</span> and <span class="math-container">$f(\sup A)$</span> are upper bounds.</p>
<p>But I don't know what else I can do.</p>
| Ross Millikan | 1,827 | <p><span class="math-container">$$s=r-\sqrt{r^2-y^2}\\
\sqrt{r^2-y^2}=s-r\\
r^2-y^2=s^2-2sr+r^2\\
2sr=s^2+y^2\\
r=\frac{s^2+y^2}{2s}$$</span>
As we squared in the third line we may have introduced an extraneous solution, so the solution needs to be checked back into the original equation</p>
|
3,692,877 | <p>Consider the following expression in three variables, <span class="math-container">$0 \leq p,s \leq 1$</span> and <span class="math-container">$n >0$</span></p>
<p><span class="math-container">$$S_{n, p, s} = \sum_{k=0}^n {n \choose k} p^k (1-p)^{n-k} e^{-s(k - np)^2}$$</span></p>
<p>If <span class="math-container">$s = 0$</span> then <span class="math-container">$S_{n, p, 0} = 1$</span>.</p>
<blockquote>
<p>Is there a closed form for the sum for <span class="math-container">$ s > 0$</span>? If not, can it be
approximated if we assume that <span class="math-container">$n$</span> is large?</p>
</blockquote>
| Oliver Díaz | 121,671 | <p>I don't think there is a closed form. As for when <span class="math-container">$n\rightarrow\infty$</span>, it seems that the limit is <span class="math-container">$0$</span> for <span class="math-container">$s>0$</span>. Here is a sketch of the proof, hoping there are no embarrassing typos.</p>
<p><span class="math-container">$S_{n,p,s}=\mathbb{E}\Big[\exp(-s(X_n-\mathbb{E}[X_n])^2\Big]$</span> where <span class="math-container">$\mathbb{E}$</span> means expectation with respect to a binomial distribution <span class="math-container">$Bi(n,p)$</span>.</p>
<p>For <span class="math-container">$n$</span> fixed, <span class="math-container">$X_n$</span> can be thought as the sum of <span class="math-container">$n$</span> i.i.d Bernoulli random variables, say <span class="math-container">$X_n\stackrel{\text{law}}{=}B_1+\ldots + B_n$</span></p>
<p>Then the expression inside expectation becomes <span class="math-container">$\exp\Big(-sp(1-p)n\big(\frac{X_n-np}{\sqrt{p(1-p)n}}\big)^2\Big)$</span></p>
<p>By Central limit theorem <span class="math-container">$\Big(\frac{X_n-np}{\sqrt{p(1-p)n}}\Big)^2$</span> converges in law to <span class="math-container">$\chi^2$</span>. There exists a coupling in which the convergence is point wise a.s. This and dominated convergence imply the convergence to <span class="math-container">$0$</span> if <span class="math-container">$s>0$</span>.</p>
<hr>
<p>Here are plot estimates of <span class="math-container">$h_n(t)=\mathbb{E}[-t(X_n-np)^2]$</span> obtained by sampling Bernoulli random variables with <span class="math-container">$p=0.5$</span>. This provides empirical evidence that indeed <span class="math-container">$h_n(t)\xrightarrow{n\rightarrow\infty} 0$</span> for <span class="math-container">$t>0$</span>. An R script is given below.</p>
<p><a href="https://i.stack.imgur.com/AZ7lm.png" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/AZ7lm.png" alt="Graph of h_n"></a></p>
<pre><code> library(latex2exp)
### Function of interest
myfunction <- function(s,x){
exp(-s*x*x)
}
### Generate point estimates at time t
myfunction.t <- function(bin.central,t){
mean(myfunction(t,bin.central))
}
Vmyfunction.t <- Vectorize(myfunction.t,vectorize.args = "t")
### Generate samples of (X-np)^2 where X~Binomial(n,p)
npop <- 10000
nsample <- 10000
p<- .5
ber.sample <- matrix(rbinom(npop*nsample,1,p),npop,nsample) - p
ber.tot.sample1 <- apply(ber.sample[1:1000,],2,sum)
ber.tot.sample2 <- apply(ber.sample[1:2500,],2,sum)
ber.tot.sample3 <- apply(ber.sample[1:5000,],2,sum)
ber.tot.sample4 <- apply(ber.sample,2,sum)
t <- seq(0,2,by = .05)
### Generate plot of point estimates of E[exp(-t*(X-np)^2)] at time t.
plot(t,Vmyfunction.t(ber.tot.sample1,t),type = "l",col = "magenta",
xlab="t", ylab="mean")
title(main=TeX("Expencted value of <span class="math-container">$\\exp(-s(X-np)^2)$</span>"),
sub=TeX("<span class="math-container">$X\\sim binom(n,0.5)$</span>"))
lines(t,Vmyfunction.t(ber.tot.sample2,t),type = "l",col = "red")
lines(t,Vmyfunction.t(ber.tot.sample3,t),type = "l",col = "green")
lines(t,Vmyfunction.t(ber.tot.sample4,t),type = "l",col = "darkgreen")
legend("topright", cex=0.6,
c("1000","2500","5000","10000"),
title="n",
fill=c("magenta","red","green","darkgreen"),
horiz=TRUE)
############ END ##########
</code></pre>
|
33,215 | <p>There is a huge debate on the internet on the value of <span class="math-container">$48\div2(9+3)$</span>.</p>
<p>I believe the answer <span class="math-container">$2$</span> as I believe it is part of the bracket operation in BEDMAS. <a href="https://www.mathway.com" rel="nofollow noreferrer">Mathway</a> yields the same answer. I also believe that if <span class="math-container">$48\div2\times(9+3)$</span> was asked it would be <span class="math-container">$288$</span> which Mathway agrees with as well.</p>
<p>However, <a href="https://www.wolframalpha.com" rel="nofollow noreferrer">WolframAlpha</a> says it is <span class="math-container">$288$</span> either way.</p>
<p>A friend of mine (who is better at math) told me that there is no such thing as 'implicit multiplication', only shorthand so that is in fact done after the division (going left to right, not necessarily because division occurs before multiplication. But he didn't explicitly give a reason)</p>
<p>What is the answer and why?</p>
| Community | -1 | <p>There is no order difference between implicit and explicit multiplications. <a href="http://www.purplemath.com/modules/orderops2.htm" rel="noreferrer">Purplemath</a> suggests that implied multiplication outside of parentheses also gets parenthetical order priority over all other multiplication(division). So they would interpret the implied multiplied parenthetical as $48\div(2\times(9+3)) = 2$.</p>
<p>Alternatively, all implicit-capable calculators I've tried give the same results as <a href="http://www.wolframalpha.com/input/?i=48%C3%B72%289%2b3%29" rel="noreferrer">Wolfram Alpha</a> which interprets the implication as $(48\div2)\cdot(9+3) = 288$</p>
<p>Some confusion also seems to be the $\div$ division symbol itself as $48/2(9+3)$ visually supports the mutual parenthetical implied multiplication of Wolfram Alpha: $\frac{48}{2}\cdot(9+3) = 288$</p>
<p>The short answer is that the formula as written is too easily misinterpreted and the author should clarify it to ensure its proper calculation.</p>
<p>PS: to further furrow your brow - employing the parenthetical as a variable yields different results. So, <a href="http://www.wolframalpha.com/input/?i=48%C3%B72c%20where%20c=%289%2b3%29" rel="noreferrer">$48\div2c$ where $c=9+3$</a> yields $2$ - but this conflates the distinction from parenthetical to coefficient-variable syntax. <a href="http://www.wolframalpha.com/input/?i=48%C3%B72*c%20where%20c=%289%2b3%29" rel="noreferrer">$48\div2\cdot c$ where $c=9+3$</a> yields $288$.</p>
|
33,215 | <p>There is a huge debate on the internet on the value of <span class="math-container">$48\div2(9+3)$</span>.</p>
<p>I believe the answer <span class="math-container">$2$</span> as I believe it is part of the bracket operation in BEDMAS. <a href="https://www.mathway.com" rel="nofollow noreferrer">Mathway</a> yields the same answer. I also believe that if <span class="math-container">$48\div2\times(9+3)$</span> was asked it would be <span class="math-container">$288$</span> which Mathway agrees with as well.</p>
<p>However, <a href="https://www.wolframalpha.com" rel="nofollow noreferrer">WolframAlpha</a> says it is <span class="math-container">$288$</span> either way.</p>
<p>A friend of mine (who is better at math) told me that there is no such thing as 'implicit multiplication', only shorthand so that is in fact done after the division (going left to right, not necessarily because division occurs before multiplication. But he didn't explicitly give a reason)</p>
<p>What is the answer and why?</p>
| Justin | 2,533 | <p>It's ambiguous, there is not one right answer in this case, other than possibly that it is undefined. You may have
$$\frac{48}{2(9+3)} = 2$$<br>
or<br>
$$\frac{48}{2}(9+3) = 288$$ </p>
<p>Therefore, there is no point in debating this. </p>
<p>Note that the reason you are get different answers from mathway and google calculator is that the algorithms they use for parsing input are different. These algorithms apparently (and understandably) leave it up to the user in this case to give input that can only be interpreted in one way. This is not the case, and is therefore why the two's answers differ.</p>
|
677,393 | <p>I have to solve an introductory counting principles problem, It goes like this:</p>
<blockquote>
<blockquote>
<p>Twelve people travel in three cars, with four people in each car. Each car is driven by its owner. Find the number of ways in which the remaining nine people may be allocated to the cars. <strong>(The arrangement within the car doesn't matter)</strong></p>
</blockquote>
</blockquote>
<p>I thought that the nine people have to be grouped into 3 cars where the arrangement doesn't matter, so the answer should be:</p>
<blockquote>
<p>9C3
= 84 </p>
</blockquote>
<p>But, this is not the right answer. So, how do i solve this?</p>
| ABcDexter | 128,946 | <p>The answer should be $(^9C_3 * ^6C_3 * ^3C_3)$ = 1680. <br>
As first we choose 3 out of 9 and then 3 Out Of remaining 6 and lastly 3 out of 3. <br></p>
<p>The mistake I did was to take extra care of the dissimilarity of the cars which caused the over-addition.</p>
<p><br>
Explanation
The correct answer is : number of ways to form first group* number of ways to form second group * third group
<br></p>
<p>Any group can occupy any car, if we permute again then it will be a mere repitition.
<br>
Multiplication rule : $(^9C_3 * ^6C_3 * ^3C_3)$ <br>
i.e.
Number of ways of filing first car * number of ways of filing second * number of ways of filing third.</p>
|
677,393 | <p>I have to solve an introductory counting principles problem, It goes like this:</p>
<blockquote>
<blockquote>
<p>Twelve people travel in three cars, with four people in each car. Each car is driven by its owner. Find the number of ways in which the remaining nine people may be allocated to the cars. <strong>(The arrangement within the car doesn't matter)</strong></p>
</blockquote>
</blockquote>
<p>I thought that the nine people have to be grouped into 3 cars where the arrangement doesn't matter, so the answer should be:</p>
<blockquote>
<p>9C3
= 84 </p>
</blockquote>
<p>But, this is not the right answer. So, how do i solve this?</p>
| Kiran | 82,744 | <p>Correct answer is ${9\choose3}\times{6\choose3}$.</p>
<p>$3!$ is not needed as it is already factored in ${9\choose3}\times{6\choose3}$. It's easy to understand if we try with small examples.</p>
|
109,569 | <p>Let's say I have a complex valued matrix $\begin{pmatrix}1+I&2+2I&3+3I\\4+4I&5+5I&6+6I\end{pmatrix}$ represented by a list:</p>
<pre><code> list = {{1 + I, 2 + 2 I, 3 + 3 I}, {4 + 4 I, 5 + 5 I, 6 + 6 I}}
</code></pre>
<p>I know how to plot each point of the matrix on the complex plane:</p>
<pre><code>Mlist = Table[Table[{Re[list[[i,j]]], Im[list[[i,j]]]}, {i,1,2}], {j,1,3}];
ListPlot[Mlist, PlotRange -> All]
</code></pre>
<p>In my case, I have 1000 rows, and I would like 2 things:</p>
<ul>
<li><p>Each point on a given row has the same color</p></li>
<li><p>The color of each row vary regularly along the number of row.</p></li>
</ul>
<p>I have no idea how to handle this. Any suggestion?</p>
| QuantumDot | 2,048 | <p>I came up with another way to do this:</p>
<pre><code>expr /. b[a_]*c_f :>
Block[{$patternMatched},
With[{result = c /. (d[a] :> ($patternMatched = True; e))},
result /; $patternMatched]]
</code></pre>
<p>Once the outer pattern matches (<code>b[a_]*c_f</code>), a boolean is set up <code>$patternMatched</code>. Then, if the inner pattern matches (<code>d[a]</code> appears inside <code>c</code>) then <code>$patternMatched</code> is set to <code>True</code>, so that the replacement is made. If the inner pattern fails to match, the <code>Condition</code> fails, stopping the outer replacement from happening.</p>
|
1,273,441 | <p>In Contests in Higher Mathematics: Miklos Schweitzer Competitions,
1962-1991 by Gabor J Szekely, problem F.57 there is the study of
$f~:~[0,\infty)\to (0,\infty)$ such that: $\exists c>0, \forall x>0$,
$f'(x)=cf(x+1)$.</p>
<p>It states, without proof, that for this equation has a solution if and only if
$c\leq 1/e$.</p>
<p>There is a reference to an unpublished paper:
T. Krisztin, Exponential bound for positive solutions of functional differential equations, unpublished manuscript</p>
<p>Does anyone have this paper (or a proof of $c\leq 1/e$)?</p>
| achille hui | 59,379 | <p>I'm going to prove the "only if" part of the problem. i.e. If $f : [0,\infty) \to (0,\infty)$ is a solution of
the equation</p>
<p>$$f'(x) = c f(x+1),\quad c > 0\tag{*1}$$</p>
<p>then $c \le \frac{1}{e}$, the other direction is leave as an exercise.</p>
<p>For any solution $f(x)$ of the problem, let $\displaystyle\;\delta = \inf\left\{ \frac{f(x+1)}{f(x)} : x > 0 \right\}\;$.</p>
<p>Notice $f'(x) = cf(x+1) > 0$ implies $f(x)$ is strictly increasing. We have $f(x+1) > f(x)$ for all $x > 0$. This means $\delta$ exists and $\ge 1$. As a result,
over the interval $(0,\infty)$, we have
$$\begin{align}
& f'(x) = cf(x+1) \ge c\delta f(x)\\
\implies & (\log f(x))' \ge c\delta\\
\implies & \log f(x+1) - \log f(x) \ge \int_x^{x+1} c\delta dx = c\delta \\
\implies & \frac{f(x+1)}{f(x)} \ge e^{c\delta}
\end{align}
$$
This leads to
$$\delta = \inf\left\{ \frac{f(x+1)}{f(x)} : x > 0 \right\} \ge e^{c\delta}
\implies c \le y e^{-y}\quad\text{ for } y = c\delta > 0$$
From this, we can conclude
$$c \le \sup\big\{ y e^{-y} : y > 0 \big\} = \frac{1}{e}$$</p>
<p>BTW, if one remove the restriction that $f(x)$ is positive, there are solutions for the delayed ODE $(*1)$ even when $c > \frac{1}{e}$.</p>
|
309,380 | <p>Let me sum up my - hopefully correct - understanding of the <a href="https://en.wikipedia.org/wiki/Travelling_salesman_problem" rel="nofollow noreferrer">travelling salesman problem</a> and <a href="https://en.wikipedia.org/wiki/Complexity_class" rel="nofollow noreferrer">complexity classes</a>. It's about <a href="https://en.wikipedia.org/wiki/Decision_problem" rel="nofollow noreferrer">decision problems</a>:</p>
<blockquote>
<p>"[...] a decision problem is a problem that can be posed as a yes-no
question of the input values. Decision problems typically appear in
mathematical questions of decidability, that is, the question of the
existence of an <strong>effective method</strong> to determine the existence of some
object."</p>
</blockquote>
<p>The travelling salesman problem (<a href="https://en.wikipedia.org/wiki/Travelling_salesman_problem" rel="nofollow noreferrer"><strong>TSP</strong></a>) - as a decision problem - is to find an answer to the question:</p>
<blockquote>
<p>Given an $n \times n$ matrix $W = (w_{ij})$ with $w_{ij} \in \mathbb{Q}$ and a number $L\in \mathbb{Q}$.</p>
<p>Is there a permutation $\pi$ of $\{1,\dots, n\}$ such that</p>
<p>$$L(\pi) = \sum_{i=1}^{n} w_{\pi(i)\pi(i+1)} < L?$$</p>
</blockquote>
<p><sub>with <a href="https://en.wikipedia.org/wiki/Modular_arithmetic" rel="nofollow noreferrer">modular</a> addition, i.e. $n+1 = 1$</sub></p>
<p>The answer can be given as a specific example (the output of a constructive "problem solver") which then can be checked for correctness. For <strong>TSP</strong> we know that a specific example given by a constructive problem solver (e.g. a specific permutation $\pi$) can be checked in polynomial time for $L(\pi) < L$, that means <strong>TSP</strong> <a href="https://en.wikipedia.org/wiki/NP_(complexity)" rel="nofollow noreferrer">$\in\mathcal{NP}$</a>. </p>
<p>But the answer may also be given by just a boolean value <strong>YES</strong> or <strong>NO</strong> , which cannot be checked at all. (What would we try to check?)</p>
<p>The first kind of answer is given by algorithms that are programmed to read arbitary matrices $W$ and numbers $L$ and give an example $\pi$. These are equivalent to constructive proofs which somehow construct a $\pi$ from given $W$ and $L$, and which may be correct or not.</p>
<p>The second kind of answer is given by non-construtive proofs - which nevertheless give an answer. Such a proof also "reads" some general $W$ and $L$ and makes some general considerations about them, e.g. like this: If numbers $x_1, \dots x_n$ can be calculated from $W$ and they relate to $L$ such that $f(x_1,\dots, x_n, L) = 0$ then the answer is <strong>YES</strong> otherwise <strong>NO</strong>.</p>
<p>My question is: </p>
<blockquote>
<p>If some day it is proved that <strong>TSP</strong> <a href="https://en.wikipedia.org/wiki/P_(complexity)" rel="nofollow noreferrer">$\not\in \mathcal{P}$</a> (because
<a href="https://en.wikipedia.org/wiki/P_versus_NP_problem" rel="nofollow noreferrer">$\mathcal{P} \neq \mathcal{NP}$</a> and <strong>TSP</strong> is <a href="https://en.wikipedia.org/wiki/NP-hardness" rel="nofollow noreferrer">$\mathcal{NP}$-hard</a>),
what do we learn about hypothetical non-construtive proofs that for given
$W$ and $L$ there exist solutions $\pi$ with $L(\pi) < L$ (<strong>YES</strong> or <strong>NO</strong>)?</p>
</blockquote>
<p>Or is the talk about such proofs only a chimera - because they are ill-defined or cannot exist for obvious reasons?</p>
<p><strong>Remark 1:</strong> Since proofs have no run-time, the things we can learn about them may concern only their length and/or complexity (in general: structure).</p>
<p><strong>Remark 2:</strong> Very short and simple algorithms may have exponential run-times.</p>
<p>To think more specifically about this: Assume there is a proof that proves:</p>
<blockquote>
<p>If you calculate numbers $x_1(W),\dots, x_m(W)$ of a quadratic matrix $W$
and you find that if $f(x_1,\dots,x_m,L) = 0$ then there is a
permutation $\pi$ with $L(\pi) < L$.</p>
</blockquote>
<p>What could be said about this (hypothetical!) proof, assuming that <a href="https://en.wikipedia.org/wiki/P_versus_NP_problem" rel="nofollow noreferrer">$\mathcal{P} \neq \mathcal{NP}$</a>?</p>
| usul | 29,697 | <p>I think the important distinction that may illuminate your question is between a proof that a particular instance $(W,L)$ belongs to the language TSP, and a proof that a particular algorithm for TSP is correct.</p>
<p>Consider the nondeterministic-Turing-machine $A$ that guesses a path $\pi$ and returns true if the length is at most $L$, returning false if all guesses fail. We can easily prove this is a correct nondeterministic algorithm for TSP. The proof is a finite fixed-size object.</p>
<p>Now consider a particular instance $(W,L)$. If we have a path $\pi$ whose distance is less than $L$, we can regard $\pi$ as a "proof" that the instance is in the TSP language. In fact, we can have a deterministic polynomial-time Turing Machine to check this proof's correctness: it simulates $A$ on the guess $\pi$ and outputs true if $A$ does. This "proof" $\pi$ will have a length that grows with the size of $(W,L)$.</p>
<p>This seems very important to your question because, again, for a given algorithm there is just a fixed finite-length proof it is correct (for all input sizes), while we can interpret nondeterministic TMs as guessing-and-checking a "proof" that a given instance is in the language.</p>
<hr>
<p>If I interpret correctly, your question supposes we have a theorem that characterizes an NP-hard language such as TSP, e.g.:</p>
<blockquote>
<p><strong>Theorem.</strong> A string $s = (W,L)$ is a valid TSP instance if and only if $f(W,L) = 0$.</p>
</blockquote>
<p>or perhaps "...if there exists a $\pi$ such that $f(W,L,\pi) = 0$."</p>
<p>We can interpret such theorems as a theorem that a certain algorithm correctly decides TSP, e.g.</p>
<blockquote>
<p><em>Alg.</em> Given $(W,L)$, return yes if $f(W,L) = 0$ and no otherwise.</p>
</blockquote>
<p>The proof of this is some finite-sized string, and I don't expect the asymptotic statement $P \neq NP$ will tell us much about that proof. But it would imply this algorithm does not run in polynomial time. So we learn about the computational complexity of checking the theorem's conditions, but seemingly not about the theorem's proof.</p>
<hr>
<p>Now, you might have asked about algorithms that are "nonconstructive" in the sense that they answer whether an instance is in the language, but they don't come with a "proof" $\pi$ of membership. We saw that nondeterministic Turing machines can be interpreted as coming with proofs that an instance is in a language. But I could have some e.g. exponential-time algorithm that checks an instance and outputs yes or no, but it doesn't really come with any short proof that the instance is in the language.</p>
<p>I don't know that $P \neq NP$ can say much about such algorithms, but you can think about more nuanced statements.
For example, if we show that $NEXP \neq NP$, then there are problems that have exponential-length proofs of membership, but not polynomial-length ones. If we show $PSPACE \supsetneq NP$, then some PSPACE-hard problem like if a chess position is winning does not generally have short proofs. Etc.</p>
|
117,500 | <p>How would you go about finding the conjugacy classes of the nonabelian group of order 21, $G:=\left\langle x,y | x^7=e=y^3, y^{-1}xy=x^2\right\rangle$?</p>
| Muhammad Ashraf Rather | 234,522 | <p>Here order of the group is $21= 3 \times 7$ . It is easy to see $3$ divides $7-1=6$. Thus, we have (upto isomorphism)two groups of order $21$. One of them is cyclic $\mathbb{Z}_{21}$ and other is non-abelian. This non-abelian is generated by two elements say $a$ and $b$ such that $|a|=3$ and $|b|=7$ and $ba=ab^r$ where $r$ is not congruent to $1$ modulo $7$ and $r^3\equiv 1 (mod 7)$. Defining conjugacy it is easy to see that there are five congugate classses and center has only the identity element. Moreover, class equation is $1+3+3+7+7\ldots$ :) </p>
|
2,292,015 | <p>The question is:
the first three terms of an arithmetic series $c_{n}$ are
$$a(1+b), a(1+3b),a(1+5b)$$
I needed to find the common difference in terms of $a$ and $b$ and then find the expression for $c_{n}$.</p>
<p>The final part I struggled with where I have to find $a$ and $b$ and the information given is
$$c_{5} = 25,c_{10} = 55$$</p>
<p>The answers for the first two parts are $difference = 2ab$ and $c_{n}=a(1+(2n-1)b)$</p>
| G Tony Jacobs | 92,129 | <p>A function from $\mathbb{R}$ to $\mathbb{R}^3$ generally looks like a path in space. In this case, the path begins at the point $(0,0,0)$ and spirals upward with an increasing radius, basically winding around an up-ward opening cone.</p>
<p>A good way to "see" such a function is to look at projections: Ignore the $z$-component, and <a href="http://www.wolframalpha.com/input/?i=plot+x%3Dt*cos(t),+y%3Dt*sin(t),+t%3D0+to+5" rel="nofollow noreferrer">you'll get a bird's-eye-view</a> Ignore the $x$- or $y$-component, and <a href="http://www.wolframalpha.com/input/?i=plot+x%3Dt*cos(t),+y%3Dt,+t%3D0+to+5" rel="nofollow noreferrer">you'll get a side view</a>.</p>
|
316,500 | <p>Let $ A $ be a commutative unital Banach algebra that is generated by a set $ Y \subseteq A $. I want to show that $ \Phi(A) $ is homeomorphic to a closed subset of the Cartesian product $ \displaystyle \prod_{y \in Y} \sigma(y) $. Moreover, if $ Y = \{ a \} $ for some $ a \in A $, I want to show that the map is onto.</p>
<p><strong>Notation:</strong> $ \Phi(A) $ is the set of characters on $ A $ and $ \sigma(y) $ is the spectrum of $ y $.</p>
<p>I tried to do this with the map
$$
f: \Phi(A) \longrightarrow \prod_{y \in Y} \sigma(y)
$$
defined by
$$
f(\phi) \stackrel{\text{def}}{=} (\phi(y))_{y \in Y}.
$$
I don’t know if $ f $ makes sense, and I can’t show that it is open or continuous. Need your help. Thank you!</p>
| Community | -1 | <p>Note that $\Phi (A)$ is compact in the w$^*$-topology. Also, $\prod \sigma(y)$ is compact Hausdorff in the product topology. For the map $f$ you defined, note that $Ker f = \{0\}$ since $Y$ generates $A$.</p>
<p>To prove continuity, take a net $\{\phi_\alpha\}_{\alpha \in I}$ in $\Phi(A)$, such that $\phi_\alpha \rightarrow \phi$ in w$^*$-topology. Then, $\phi_\alpha (y) \rightarrow \phi(y)$ in norm topology, for any $y \in Y$. ------------- (*)</p>
<p>Now consider a basic open set $V$ around the point $\prod_{y \in Y} \phi(y)$. Then, there exists $y_1, y_2, \ldots, y_k \in Y$ such that $V = \prod_{y \in Y} V_y $, where $V_y = \sigma (y)$ for any $y \in Y \setminus \{y_1, y_2, \ldots, y_k\}$ and $ V_{y_i} = V_i $ is an open ball in $\sigma (y_i)$ containing $\phi(y_i)$, for $i = 1, 2, \ldots, k$.</p>
<p>Using (*) we get that for each $i$, $\exists$ $\alpha_i$ such that $\phi_{\beta} (y_i) \in V_i$ for any $\beta \geq \alpha_i$. Since the index set $I$ is directed, $\exists$ $\alpha_0 \in I$ such that $\alpha_0 \geq \alpha_i$ for all $i$. Thus for any $\beta \geq \alpha_0$, we have that $\phi_{\beta} (y_i) \in V_i$ for each $i$, and hence $\prod_{y \in Y} \phi_{\beta} (y) \in V$. </p>
<p>Thus, it follows that $\prod \phi_\alpha (y) \rightarrow \prod \phi(y)$ in $\prod \sigma(y)$, i.e, $f$ is continuous.</p>
|
316,500 | <p>Let $ A $ be a commutative unital Banach algebra that is generated by a set $ Y \subseteq A $. I want to show that $ \Phi(A) $ is homeomorphic to a closed subset of the Cartesian product $ \displaystyle \prod_{y \in Y} \sigma(y) $. Moreover, if $ Y = \{ a \} $ for some $ a \in A $, I want to show that the map is onto.</p>
<p><strong>Notation:</strong> $ \Phi(A) $ is the set of characters on $ A $ and $ \sigma(y) $ is the spectrum of $ y $.</p>
<p>I tried to do this with the map
$$
f: \Phi(A) \longrightarrow \prod_{y \in Y} \sigma(y)
$$
defined by
$$
f(\phi) \stackrel{\text{def}}{=} (\phi(y))_{y \in Y}.
$$
I don’t know if $ f $ makes sense, and I can’t show that it is open or continuous. Need your help. Thank you!</p>
| Haskell Curry | 39,362 | <p><strong>The mapping $ f $ is well-defined.</strong></p>
<p>Given each $ \phi \in \Phi(A) $, we must have $ \phi(a) \in \sigma(a) $ for all $ a \in A $. Indeed, as $ \phi(a - \phi(a) \cdot \mathbf{1}_{A}) = 0 $, we see that $ a - \phi(a) \cdot \mathbf{1}_{A} $ is not invertible, or equivalently, $ \phi(a) \in \sigma(a) $. Therefore, the mapping
$$
f: \Phi(A) \longrightarrow \prod_{y \in Y} \sigma(y)
$$
as defined above makes sense.</p>
<hr>
<p><strong>The mapping $ f $ is continuous.</strong></p>
<p>Equip $ \Phi(A) $ with the weak$ ^{*} $-topology inherited from $ A^{*} $. To prove that $ \displaystyle f: \Phi(A) \to \prod_{y \in Y} \sigma(y) $ is continuous, it suffices to show that $ p_{y_{0}} \circ f: \Phi(A) \to \mathbb{R} $ is continuous for each $ y_{0} \in Y $, where $ p_{y_{0}} $ is the projection mapping of $ \displaystyle \prod_{y \in Y} \sigma(y) $ onto the $ y_{0} $-th coordinate. As
$$
\forall \phi \in \Phi(A): \quad \left( p_{y_{0}} \circ f \right)(\phi) = \phi(y_{0}),
$$
we see that $ p_{y_{0}} \circ f $ is simply the mapping $ \phi \longmapsto \phi(y_{0}) $, which is obviously continuous with respect to the weak$ ^{*} $-topology on $ \Phi(A) $. Therefore, as $ y_{0} \in Y $ is arbitrary, it follows that $ f $ is continuous.</p>
<hr>
<p><strong>The mapping $ f $ is injective.</strong></p>
<p>Observe that $ f $ is injective because any $ \phi \in \Phi(A) $ is uniquely determined by its values on a generating set for $ A $.</p>
<hr>
<p><strong>The mapping $ f $ is a topological embedding.</strong></p>
<p>To show that $ \displaystyle f: \Phi(A) \to \prod_{y \in Y} \sigma(y) $ is a topological embedding, observe firstly that $ \Phi(A) $ is weak$ ^{*} $-compact (as it is a weak$ ^{*} $-closed subset of $ \text{Ball}(A^{*}) $, which is weak$ ^{*} $-compact by the Banach-Alaoglu Theorem) and $ \displaystyle \prod_{y \in Y} \sigma(y) $ is Hausdorff. Then use the fact that <a href="http://topospaces.subwiki.org/wiki/Injection_from_compact_to_Hausdorff_implies_embedding" rel="nofollow">a continuous injection from a compact space to a Hausdorff space is a topological embedding</a>.</p>
<hr>
<p><strong>If $ A $ is generated by a single element, then $ f $ is onto.</strong></p>
<p>Suppose that $ Y = \{ y_{0} \} $. Then
$$
\sigma(y_{0}) = \{ \phi(y_{0}) ~|~ \phi \in \Phi(A) \},
$$
which shows that $ f: \Phi(A) \to \sigma(y_{0}) $ is onto.</p>
|
514,338 | <p>Okay so my algebra knowledge is pretty guff..</p>
<p>I am taking a control systems class and pretty much all the questions I am expected to revise, are about doing this algebraic manipulation and I don't know what steps the tutor is taking to do it..</p>
<p>Okay here goes..</p>
<p>If the transfer function of a system is $G(s) = 3/(20s+1)$, then the closed loop version of that is</p>
<p>$$G(s)/(G(s) + 1)$$ </p>
<p>so that would be </p>
<p>$$\frac{\frac{3}{20s+1}}{\frac{3}{20s+1} + 1}$$</p>
<p>This is the bit I am having trouble with.. he just then cancels it all out and gives us the answer on the next line which is.. $$\frac{3}{20s+4}$$</p>
<p>He gives us loads of problems to this which are all similar but I just cannot work them out as my algebra sucks so bad.. I don't know how to cancel out stuff which has a division but with an addition in the denominator..</p>
<p>I have taken a screen shot of the pdf here
<a href="https://i.imgur.com/t82sfYr.png" rel="nofollow noreferrer">http://i.imgur.com/t82sfYr.png</a></p>
<p><a href="https://i.imgur.com/GoCBjuq.png" rel="nofollow noreferrer">And another one of another pdf which explains nearly how to do it but misses out the steps..</a></p>
| Mark Bennet | 2,906 | <p>Note that $$\frac {\frac 3{20s+1}}{\frac3{20s+1}+1}=\frac {\frac 3{20s+1}}{\frac3{20s+1}+1}\cdot\frac {20s+1}{20s+1}=\frac{3}{3+20s+1}=\frac 3{20s+4}$$</p>
|
514,338 | <p>Okay so my algebra knowledge is pretty guff..</p>
<p>I am taking a control systems class and pretty much all the questions I am expected to revise, are about doing this algebraic manipulation and I don't know what steps the tutor is taking to do it..</p>
<p>Okay here goes..</p>
<p>If the transfer function of a system is $G(s) = 3/(20s+1)$, then the closed loop version of that is</p>
<p>$$G(s)/(G(s) + 1)$$ </p>
<p>so that would be </p>
<p>$$\frac{\frac{3}{20s+1}}{\frac{3}{20s+1} + 1}$$</p>
<p>This is the bit I am having trouble with.. he just then cancels it all out and gives us the answer on the next line which is.. $$\frac{3}{20s+4}$$</p>
<p>He gives us loads of problems to this which are all similar but I just cannot work them out as my algebra sucks so bad.. I don't know how to cancel out stuff which has a division but with an addition in the denominator..</p>
<p>I have taken a screen shot of the pdf here
<a href="https://i.imgur.com/t82sfYr.png" rel="nofollow noreferrer">http://i.imgur.com/t82sfYr.png</a></p>
<p><a href="https://i.imgur.com/GoCBjuq.png" rel="nofollow noreferrer">And another one of another pdf which explains nearly how to do it but misses out the steps..</a></p>
| user91011 | 91,011 | <p>\begin{align}
\frac{\frac{3}{20s+1}}{\frac{3}{20s+1}+1} &= \frac{{20s+1}}{{20s+1}}\frac{\frac{3}{20s+1}}{\frac{3}{20s+1}+1}\\
& = \frac{\frac{3(20s+1)}{20s+1}}{\frac{3(20s+1)}{20s+1}+(20s+1)}\\
& = \frac{3}{3+(20s+1)}.
\end{align}</p>
|
1,295,259 | <p>How to prove that;</p>
<blockquote>
<p>$a^{|G|}=e$ if a $\in G $</p>
</blockquote>
<p>if $G$ is a finite group and $e$ is its identity.</p>
<p>I think this could be done through pigeonhole principle but I don't want to use the Lagrange theorem.</p>
<p>How should I start?</p>
| Dietrich Burde | 83,966 | <p>I only know a proof without Lagrange (or Lagrange in disguise) for an abelian group. Suppose that $G=\{ a_1,\ldots ,a_n\}$, and set $g:=a_1a_2\cdots a_n$. Then for every $x\in G$ the map $a_i\mapsto xa_i$ is a permutation of $G$, so that
$$
g=(xa_1)(xa_2)\cdots (xa_n)=x^na_1a_2\cdots a_n=x^ng.
$$
This implies $x^{\mid G\mid}=x^n=e$.</p>
|
1,041,731 | <p>I want to prove that if $A$ in an infinite set, then the cartesian product of $A$ with 2 (the set whose only elements are 0 and 1) is equipotent to $A$.</p>
<p>I'm allowed to use Zorn's Lemma, but I can't use anything about cardinal numbers or cardinal arithmetic (since we haven't sotten to that topic in the course).</p>
<p>I read a proof of the fact that if $a$ is an infinite cardinal number, then $a+a=a$, which is something similar to what I want to prove. </p>
<p>Any suggestions will be appreciated :)</p>
| Rodrigo de Azevedo | 339,790 | <p>Visual inspection tells us that matrix <span class="math-container">$\rm A$</span> is a <a href="http://en.wikipedia.org/wiki/Companion_matrix" rel="nofollow noreferrer">companion matrix</a> and that <span class="math-container">$1$</span> is an eigenvalue of <span class="math-container">$\rm A$</span>. Hence, the characteristic polynomial of <span class="math-container">$\rm A$</span> is</p>
<p><span class="math-container">$$q (s) := s^3 - 4 s^2 + 5 s - 2 = (s - 1) (s^2 - 3 s + 2) = \color{blue}{(s-1)^2 (s-2)}$$</span></p>
|
83,965 | <p>When students learn multivariable calculus they're typically barraged with a collection of examples of the type "given surface X with boundary curve Y, evaluate the line integral of a vector field Y by evaluating the surface integral of the curl of the vector field over the surface X" or vice versa. The trouble is that the vector fields, curves and surfaces are pretty much arbitrary except for being chosen so that one or both of the integrals are computationally tractable.</p>
<p>One more interesting application of the classical Stokes theorem is that it allows one to interpret the curl of a vector field as a measure of swirling about an axis. But aside from that I'm unable to recall any other applications which are especially surprising, deep or interesting. </p>
<p>I would like use Stokes theorem show my multivariable calculus students something that they enjoyable. Any suggestions?</p>
| J.C. Ottem | 3,996 | <p>The <a href="http://www.math.duke.edu/~wka/math204/fixed.pdf">proof of Brower's fixed point theorem</a> using Stokes' theorem is a nice application I think.</p>
|
76,360 | <p>I have a string of characters, like</p>
<pre><code>"CDABOZPVRYXSWQEGNILUTHMKJF"
</code></pre>
<p>and want to convert it to a string in which the character at position <code>p</code> is bold. </p>
<p>After doing this I want to leave the result in a table that will be displayed in <code>TableForm</code> and then used for publications or export — e.g., copy as <em>LaTeX</em> — so the result needs to be (essentially) a formatted string with a single bold character.</p>
<p>Is there a way to accomplish this?</p>
| Mr.Wizard | 121 | <p>This question is related to at least:</p>
<ul>
<li><a href="https://mathematica.stackexchange.com/q/7732/121">Highlighting text with StringReplacePart but also using Style, Subscript</a></li>
<li><a href="https://mathematica.stackexchange.com/q/10990/121">How to join two Style[]d strings</a></li>
</ul>
<p>Fortunately it is simpler than the first one and we can apply the methods provided in the second one.</p>
<pre><code>stringBold[s_String, pos_] :=
"" <> MapAt[Style[#, Bold] ~ToString~ StandardForm &, Characters@s, pos]
stringBold["CDABOZPVRYXSWQEGNILUTHMKJF", 7]
</code></pre>
<p><img src="https://i.stack.imgur.com/s2ctV.png" alt="enter image description here"></p>
<p>The output is a <code>String</code> with and embedded Box form:</p>
<pre><code>% // InputForm
</code></pre>
<blockquote>
<pre><code>"CDABOZ\!\(\*StyleBox[\"\\\"P\\\"\", Bold, Rule[StripOnInput, \
False]]\)VRYXSWQEGNILUTHMKJF"
</code></pre>
</blockquote>
<p>Any specification that <code>MapAt</code> accepts can be used for parameter <code>pos</code>:</p>
<pre><code>stringBold["CDABOZPVRYXSWQEGNILUTHMKJF", 2 ;; -3 ;; 3]
</code></pre>
<p><img src="https://i.stack.imgur.com/A4So7.png" alt="enter image description here"></p>
<p>However the function <em>cannot</em> be applied to more than once, or to an already styled string:</p>
<pre><code>stringBold[%, 3] (* failure *)
</code></pre>
|
76,360 | <p>I have a string of characters, like</p>
<pre><code>"CDABOZPVRYXSWQEGNILUTHMKJF"
</code></pre>
<p>and want to convert it to a string in which the character at position <code>p</code> is bold. </p>
<p>After doing this I want to leave the result in a table that will be displayed in <code>TableForm</code> and then used for publications or export — e.g., copy as <em>LaTeX</em> — so the result needs to be (essentially) a formatted string with a single bold character.</p>
<p>Is there a way to accomplish this?</p>
| kglr | 125 | <pre><code>srF = StringReplacePart[#, ToString[Style[StringTake[#, {#2}], ##3], StandardForm], {#2, #2}] &;
str = "CDABOZPVRYXSWQEGNILUTHMKJF";
srF[str, #, Red, Bold, 16] & /@ {3,9}
</code></pre>
<p><img src="https://i.stack.imgur.com/mwEAu.png" alt="enter image description here"></p>
<p>Note: Copy as LateX does not work</p>
|
3,999,699 | <p>I want to show that <span class="math-container">$a_{n} = \sqrt{n}$</span> is not a bounded sequence.</p>
<p>Definition: We say that a sequence is bounded if it is bounded above and below. A sequence <span class="math-container">$a_{n}$</span> is bounded above if there exists <span class="math-container">$C$</span> such that, for all <span class="math-container">$n$</span>, <span class="math-container">$a_{n} \leq C$</span>. A sequence <span class="math-container">$a_{n}$</span> is bounded below if there exists <span class="math-container">$C$</span> such that, for all <span class="math-container">$n$</span>, <span class="math-container">$a_{n} \geq C$</span>.</p>
<p>Well clearly, our sequence is bounded below by <span class="math-container">$0$</span>. So we need to show it is not bounded above.</p>
<p>Proof:</p>
<p>We need to show that for every <span class="math-container">$C$</span>, there exists an <span class="math-container">$n$</span> such that <span class="math-container">$a_{n} > C$</span>. Observing that for every <span class="math-container">$C \geq 0, \sqrt{C^{2}} = C$</span>. Suppose that <span class="math-container">$ n > C^{2}$</span> then we have <span class="math-container">$\sqrt{n} > C$</span></p>
| Ted Shifrin | 71,348 | <p>The sloppy reasoning occurs with the right triangle you wrote. You must be careful about the <em>range</em> (values) of the arcsec function. <span class="math-container">$y=\text{arcsec}(x)$</span> can lie <em>either</em> in <span class="math-container">$[0,\pi/2)$</span> <em>or</em> in <span class="math-container">$(\pi/2,\pi]$</span>. (These correspond, respectively, to <span class="math-container">$x>0$</span> and <span class="math-container">$x<0$</span>.)</p>
<p><span class="math-container">$\tan(y)<0$</span> when <span class="math-container">$y\in (\pi/2,\pi]$</span>, and so <span class="math-container">$\tan(y)=-\sqrt{x^2-1}$</span> in that event. Working with your formula for <span class="math-container">$y'$</span>, we note that when <span class="math-container">$x<0$</span>,
<span class="math-container">$$\sec(y)\tan(y)= x(-\sqrt{x^2-1})= (-x)\sqrt{x^2-1} =|x|\sqrt{x^2-1},$$</span>
and this explains the formula. To be honest, it's a bit sneaky to move the negative to the other term, but it allows us to write a single formula, rather than writing down cases. (There's no problem, of course, when <span class="math-container">$x>0$</span>.)</p>
|
1,083,277 | <p>$a,b,c \in \mathbb{R}$ and $a+b+c=0$.
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$</p>
<p>I think that $2^{a}.2^{b}.2^{c}=1$, but i don't know what to do next</p>
| DeepSea | 101,504 | <p><strong>Hint:</strong> $f(x) = x^3 - x$, $g(x) = \dfrac{x^3}{9} - x$ $\text{ increases}$, $x \geq 3$ and Jensen's inequality !</p>
|
1,083,277 | <p>$a,b,c \in \mathbb{R}$ and $a+b+c=0$.
Prove that: $8^{a}+8^{b}+8^{c}\geqslant 2^{a}+2^{b}+2^{c}$</p>
<p>I think that $2^{a}.2^{b}.2^{c}=1$, but i don't know what to do next</p>
| math110 | 58,742 | <p>Let $x=2^a,y=2^b,z=2^c$. Then $x,y,z>0$ and
$$ xyz=2^{a+b+c}=1.$$
By Holder's inequality
$$(x^3+y^3+z^3)(1+1+1)(1+1+1)\ge (x+y+z)^3.$$
Therefore,
$$x^3+y^3+z^3\ge\dfrac{(x+y+z)^3}{9}\ge (x+y+z)$$
because use AM-GM inequality
$$(x+y+z)^2\ge (3\sqrt[3]{xyz})^2=9.$$</p>
|
3,364,404 | <blockquote>
<p>If <span class="math-container">$X\sim \operatorname{Binom}(n,p)$</span> and <span class="math-container">$Y\sim \operatorname{Ber}\left(\frac Xn\right)$</span>, then find <span class="math-container">$E[X\mid Y]$</span>.</p>
</blockquote>
<p>Is there a name for such a random variable <span class="math-container">$Y$</span>, where its distribution depends on another r.v. ?</p>
<p>I got after some lengthy calculation <span class="math-container">$$P(Y=0)=1-p$$</span>
<span class="math-container">$$P(Y=1)=p$$</span></p>
<p>I don't know over which variable to sum ? I must get a function in terms of <span class="math-container">$Y$</span> </p>
<p>Then <span class="math-container">$$E[X\mid Y=1]=\sum_\limits{???}\frac{P(X=k)P(Y=1\mid X=k)}{P(Y=1)}$$</span></p>
<p>what I know, (assuming there are <span class="math-container">$n$</span> trials with <span class="math-container">$n\ge k$</span>)</p>
<p><span class="math-container">$$P(X=k)=\binom{n}{k}p^k(1-p)^{n-k}$$</span></p>
<p><span class="math-container">$$P(Y=1\mid X=k)=\frac{k}{n}$$</span></p>
<p>It is also inappropriate to set <span class="math-container">$Y=1$</span> instead of <span class="math-container">$Y=m$</span> for some <span class="math-container">$m$</span>, otherwise I cannot recognize <span class="math-container">$Y$</span> in the formula. How to remedy this ?</p>
<p>EDIT: When the variable goes from <span class="math-container">$k=0$</span> to <span class="math-container">$n$</span> I obtain the following,</p>
<p><span class="math-container">$$E[X\mid Y=1]=np-p+1,\quad E[X\mid Y=0]=p(n-1)$$</span></p>
<p>so it seems legitimate to combine these 2 in the following way,</p>
<p><span class="math-container">$$E[X\mid Y]=(np-p+1)Y+\left(p(n-1)\right)(1-Y)$$</span></p>
<p>Is this true ?</p>
| drhab | 75,923 | <p>In this answer for convenience <span class="math-container">$q:=1-p$</span> and <span class="math-container">$Z\sim\mathsf{Bin}\left(n-1,p\right)$</span>.</p>
<p>Working out: <span class="math-container">$$P\left(Y=1\right)=\sum_{k=0}^{n}P\left(Y=1\mid X=k\right)P\left(X=k\right)$$</span>
we find <span class="math-container">$P\left(Y=1\right)=p$</span> and consequently <span class="math-container">$P\left(Y=0\right)=q$</span>.</p>
<p>Then for <span class="math-container">$k=1,\dots,n$</span>:</p>
<p><span class="math-container">$$P\left(X=k\mid Y=1\right)p=P\left(X=k\wedge Y=1\right)=P\left(Y=1\mid X=k\right)P\left(X=k\right)=\frac{k}{n}\binom{n}{k}p^{k}q^{n-k}$$</span>
leading to: <span class="math-container">$$P\left(X=k\mid Y=1\right)=\binom{n-1}{k-1}p^{k-1}q^{n-k}$$</span></p>
<p>This reveals that <span class="math-container">$\left(X-1\mid Y=1\right)\stackrel{d}{=}Z$</span> so that
<span class="math-container">$$\mathbb{E}\left[X\mid Y=1\right]=1+\mathbb{E}\left[X-1\mid Y=1\right]=1+\mathbb{E}Z=1+\left(n-1\right)p\tag1$$</span></p>
<p>Similarly we find for <span class="math-container">$k=0,\dots,n-1$</span>:<span class="math-container">$$P\left(X=k\mid Y=0\right)q=P\left(X=k\wedge Y=0\right)=P\left(Y=0\mid X=k\right)P\left(X=k\right)=\left(1-\frac{k}{n}\right)\binom{n}{k}p^{k}q^{n-k}$$</span>
so that <span class="math-container">$$P\left(X=k\mid Y=0\right)=\binom{n-1}{k}p^{k}q^{n-1-k}$$</span></p>
<p>This reveals that <span class="math-container">$\left(X\mid Y=0\right)\stackrel{d}{=}Z$</span> so that
<span class="math-container">$$\mathbb{E}\left[X\mid Y=0\right]=\mathbb{E}Z=\left(n-1\right)p\tag2$$</span></p>
<p>Based on <span class="math-container">$(1)$</span> and <span class="math-container">$(2)$</span> we conclude: <span class="math-container">$$\mathbb{E}\left[X\mid Y\right]=Y+\left(n-1\right)p$$</span></p>
<hr>
<p><strong>Edit</strong>: </p>
<p>More simple approach that arose later and was inspired by the outcome <span class="math-container">$\mathbb E[X\mid Y]=Y+(n-1)p$</span>.</p>
<p>Let us set <span class="math-container">$X=Y+Z$</span> where <span class="math-container">$X,Y$</span> are independent with <span class="math-container">$Y\sim\mathsf{Bern}\left(p\right)$</span>
and <span class="math-container">$Z\sim\mathsf{Bin}\left(n-1,p\right)$</span>. </p>
<p>Then observe that <span class="math-container">$X\sim\mathsf{Bin}\left(n,p\right)$</span> and that <span class="math-container">$\left(Y\mid X\right)\sim\mathsf{Bern}\left(\frac{X}{n}\right)$</span>.</p>
<p>This is exactly the situation prescribed in your question except that
you stated that <span class="math-container">$Y\sim\mathsf{Bern}\left(\frac{X}{n}\right)$</span>.</p>
<p>That is actually not correct as <strong>Graham Kemp</strong> made clear in his answer.</p>
<p>In this situation we find directly: <span class="math-container">$$\mathbb{E}\left[X\mid Y\right]=\mathbb{E}\left[Y+Z\mid Y\right]=\mathbb{E}\left[Y\mid Y\right]+\mathbb{E}\left[Z\mid Y\right]=Y+\mathbb{E}Z=Y+\left(n-1\right)p$$</span>where the third equality is based on independence.</p>
|
650,395 | <p>I am given generating functions $f(x)= \frac{x}{1-x}$ or $f(x)=\frac{1}{1+x^{2}}$ or $f(x)=\frac{1}{x^2-5x+6}$ and I am obliged to write sequence which are generated by this functions. What is the fastest algorithm to solve these problems? I have problem with even starting. I will be glad if anyone would be so nice to explain me algorithm to solve this kind of exercises or post any reference that is related with my problem.</p>
| Adi Dani | 12,848 | <p>$$\frac{x}{1-x}=x\frac{1}{1-x}=x\sum_{n=0}^{\infty}x^n=\sum_{n=0}^{\infty}x^{n+1}=\sum_{n=1}^{\infty}x^{n}$$
$$\frac{1}{1+x^2}=\frac{1}{1-(-x^2)}=\sum_{n=0}^{\infty}(-x^2)^n=\sum_{n=0}^{\infty}(-1)^nx^{2n}$$</p>
|
2,143,510 | <p>NOTE: There are some other similar questions, but I got a negative answer to this question from my proof. Please find out the errors in my reasoning. </p>
<p>$\mathbf {Claim:}$ Is every point of every open set $E \subset R^2$ a limit point of E? Answer the same question for closed sets in $E \subset R^2$</p>
<p>From "Baby Rudin"</p>
<p>$\mathbf {Proof:}$ $\emptyset$ is both open and closed in every topological space. $R^2$ is a metric space, which is a kind of topological space, so $\emptyset$ is both open and closed in it. $\emptyset$ has no limit point because its neighborhood has no other point to include, so we can get a negative answer to both questions.</p>
<p>If possible, please have a look at the two questions that I got while reading the answer to this similiar question:
<a href="https://math.stackexchange.com/questions/1446589/proof-that-every-point-of-every-open-set-e">Proof that every point of every open set E⊂ℝ^2 is a limit point of E?</a>⊂ℝ2-is-a-limit-point-of-e</p>
<ol>
<li>"$q_s=(x_1+s,x_2)$" should be $q_s=(x_1+s,y_1)$, right?</li>
<li>I still don't understand why there should be $\epsilon$. Why can't r complete the proof?</li>
</ol>
<p>I don't have the right to comment on the original post led by the above link, so I ask the two questions here.
Finally, I find this forum very active, responsive and helpful, but not quite friendly to newcomers.</p>
| John | 105,625 | <p>For the open set part, you are asked to show whether the following statement is correct:</p>
<p>Let $E$ be an open set in $\mathbb{R}^2$. If $p\in E$, then $p$ is a limit point of $E$.</p>
<p>Your counterexample by letting $E=\emptyset$ is not correct. Because by definition, empty set contains no element, which means you can't select element from it, hence the if part of the above statement is false. Then for the statement $P\rightarrow Q$, if $P$ is false, no matter whether $Q$ is true or false, we say $P\rightarrow Q$ is true. Therefore, for $E=\emptyset$, we know the statement is true.</p>
<p>For the proof in your link:</p>
<ol>
<li><p>Yes. $q_s=(x_1+s,y_1)$.</p></li>
<li><p>We are asked to show $p$ is a limit point of $E$, which means for any $\epsilon>0$, we need to find a $q_\epsilon\in E\cap B(p,\epsilon)\setminus\{p\}$.</p></li>
</ol>
|
2,143,510 | <p>NOTE: There are some other similar questions, but I got a negative answer to this question from my proof. Please find out the errors in my reasoning. </p>
<p>$\mathbf {Claim:}$ Is every point of every open set $E \subset R^2$ a limit point of E? Answer the same question for closed sets in $E \subset R^2$</p>
<p>From "Baby Rudin"</p>
<p>$\mathbf {Proof:}$ $\emptyset$ is both open and closed in every topological space. $R^2$ is a metric space, which is a kind of topological space, so $\emptyset$ is both open and closed in it. $\emptyset$ has no limit point because its neighborhood has no other point to include, so we can get a negative answer to both questions.</p>
<p>If possible, please have a look at the two questions that I got while reading the answer to this similiar question:
<a href="https://math.stackexchange.com/questions/1446589/proof-that-every-point-of-every-open-set-e">Proof that every point of every open set E⊂ℝ^2 is a limit point of E?</a>⊂ℝ2-is-a-limit-point-of-e</p>
<ol>
<li>"$q_s=(x_1+s,x_2)$" should be $q_s=(x_1+s,y_1)$, right?</li>
<li>I still don't understand why there should be $\epsilon$. Why can't r complete the proof?</li>
</ol>
<p>I don't have the right to comment on the original post led by the above link, so I ask the two questions here.
Finally, I find this forum very active, responsive and helpful, but not quite friendly to newcomers.</p>
| DanielWainfleet | 254,665 | <p>You can do this more easily by noting that $\mathbb R^2$ has no isolated points. A point $p$ in a space $X$ is isolated iff $\{p\}$ is open. </p>
<p>Let $p\in U\subset R^2$ where $U$ is open. If $V$ is any nbhd of $p,$ then there exists open $V'$ with $p\in V'\subset V.$ </p>
<p>Now $U\cap V'$ is open and not empty (as it contains $p$) so $\phi \ne U\cap V'\ne \{p\}.$ So there exists $q\in U\cap V'$ with $q\ne p.$</p>
<p>So any nbhd $V$ of $p$ contains a point $q$ with $p\ne q\in U.$ </p>
|
2,084,624 | <p>The question is : </p>
<p>Is $\sum_{k=1}^\infty \frac{(-3)^k(k!)}{k^k}$ convergent? </p>
<p>Note : I can't find the limit of its main term. I know the answer must be related to some test about convergence of series ... I don't know which one and i can't find the limit.</p>
| Behrouz Maleki | 343,616 | <p>Apply <a href="https://en.wikipedia.org/wiki/Stirling's_approximation" rel="nofollow noreferrer">Stirling's approximation</a>
$$k!\sim \sqrt{2k\pi}\left(\frac{k}{e}\right)^k$$
and use <a href="https://en.wikipedia.org/wiki/Root_test" rel="nofollow noreferrer">Root Test</a>.</p>
|
148,807 | <p>I'm not sure if these types of questions are accepted here or not (I'm very sorry if it's not), but it would be great if anyone could explain me this.</p>
<blockquote>
<p><strong>Question:</strong>
Using his bike, Daniel can complete a paper route in 20 minutes. Francisco, who walks the route, can complete it in 30 minutes. How long will it take the two boys to complete the route if they work together, one starting at each end of the route?</p>
</blockquote>
<p>I have the answer: 12 minutes</p>
<p>But I don't understand the solution given in the book.</p>
<p>Can any of you explain how to solve this? Your help is highly appreciated.</p>
| Inquest | 35,001 | <p>Daniel finishes $\frac{1}{20}$ th of his work in 1 min. Fransico finishes $\frac{1}{30}$ th of his work in 1 min. In one min (simultaneously), they finish off $\frac{1}{30}$ + $\frac{1}{20}$ = $\frac{1}{12}$ of the work (Assuming no dependency which is true in this case as they are starting from opposite ends). So, in one minute, they finish off $\frac{1}{12}$ th of the work. So, in 12 minutes, they will finish off the entire work.</p>
|
3,397,548 | <p>For a sequence <span class="math-container">$\{x_n\}_{n=1}^{\infty}$</span>, define <span class="math-container">$$\Delta x_n:=x_{n+1}-x_n,~\Delta^2 x_n:=\Delta x_{n+1}-\Delta x_n,~(n=1,2,\ldots)$$</span> which are named <strong>1-order</strong> and <strong>2-order difference</strong>, respectively. </p>
<p>The problem is stated as follows:</p>
<blockquote>
<p>Let <span class="math-container">$\{x_n\}_{n=1}^{\infty}$</span> be <strong>bounded</strong> , and satisfy
<span class="math-container">$\lim\limits_{n \to \infty}\Delta^2 x_n=0$</span>. Prove or disprove
<span class="math-container">$\lim\limits_{n \to \infty}\Delta x_n=0.$</span></p>
</blockquote>
<p>By intuiton, the conclusion is likely to be true. According to <span class="math-container">$\lim\limits_{n \to \infty}\Delta^2 x_n=0,$</span> we can estimate <span class="math-container">$\Delta x_n$</span> almost equal with an increasing <span class="math-container">$n$</span>. Thus, <span class="math-container">$\{x_n\}$</span> looks like an <strong>arithmetic sequence</strong>. If <span class="math-container">$\lim\limits_{n \to \infty}\Delta x_n \neq 0$</span>, then <span class="math-container">$\{x_n\}$</span> can not be bounded.</p>
<p>But how to prove it rigidly?</p>
| Martin R | 42,969 | <p>Yes: If <span class="math-container">$(x_n)$</span> is bounded and <span class="math-container">$\lim_{n \to \infty}\Delta^2 x_n = 0$</span> then <span class="math-container">$\lim_{n \to \infty}\Delta x_n = 0$</span>. That is a consequence of the following general estimate:</p>
<blockquote>
<p>If <span class="math-container">$(x_n)$</span> is a sequence with <span class="math-container">$|x_n| \le M$</span> and <span class="math-container">$|\Delta^2 x_n| \le K$</span> for all <span class="math-container">$n$</span> then
<span class="math-container">$$ \tag{*}
|\Delta x_n|^2 \le 4MK \, .
$$</span> for all <span class="math-container">$n$</span>.</p>
</blockquote>
<p>In our case <span class="math-container">$\lim_{n \to \infty}\Delta^2 x_n=0$</span>, so that the above can be applied to tail sequences <span class="math-container">$(x_n)_{n \ge n_0}$</span> with <span class="math-container">$K$</span> arbitrarily small, and <span class="math-container">$\lim_{n \to \infty}\Delta x_n=0$</span> follows. </p>
<p><em>Proof</em> of the claim. It suffices to prove <span class="math-container">$(*)$</span> for <span class="math-container">$n=0$</span>. Without loss of generality assume that <span class="math-container">$\Delta x_0 \ge 0$</span>. We have
<span class="math-container">$$
x_n = x_0 + \sum_{j=0}^{n-1} \Delta x_j
= x_0 + \sum_{j=0}^{n-1} \left( \Delta x_0 + \sum_{k=0}^{j-1} \Delta^2 x_k \right) \\
= x_0 + n \Delta x_0 + \sum_{j=0}^{n-1}\sum_{k=0}^{j-1} \Delta^2 x_k \, .
$$</span>
Using the given bounds <span class="math-container">$-M \le x_n \le M$</span> and <span class="math-container">$\Delta^2 x_n \ge -K$</span> it follows that
<span class="math-container">$$
M \ge -M + n \Delta x_0 - \frac{(n-1)n}{2}K \\
\implies 0 \le \frac{(n-1)n}{2}K - n \Delta x_0 + 2M
$$</span></p>
<p>If <span class="math-container">$K=0$</span> then <span class="math-container">$0 \le \Delta x_0 \le 2M/n$</span> implies <span class="math-container">$\Delta x_0 = 0$</span>, and we are done. Otherwise the quadratic inequality can be rearranged (by “completing the square”) to
<span class="math-container">$$
0 \le \left(n - \left(\frac{\Delta x_0}{K} + \frac 12 \right) \right)^2
+ \frac{4M}{K} - \left(\frac{\Delta x_0}{K} + \frac 12 \right)^2 \, .
$$</span></p>
<p>Now choose the non-negative integer <span class="math-container">$n$</span> such <span class="math-container">$\left| n - \left(\frac{\Delta x_0}{K} + \frac 12 \right) \right| \le \frac 12$</span>. Then
<span class="math-container">$$
0 \le \frac 14 + \frac{4M}{K} - \left(\frac{\Delta x_0}{K} + \frac 12 \right)^2 = \frac{4M}{K} - \left(\frac{\Delta x_0}{K} \right)^2 - \frac{\Delta x_0}{K} \\
\le \frac{4M}{K} - \left(\frac{\Delta x_0}{K} \right)^2
$$</span>
and the desired conclusion <span class="math-container">$(*)$</span> follows.</p>
<hr>
<p><em>Remarks:</em> There is a “similar” inequality for differentiable functions: </p>
<blockquote>
<p>Let <span class="math-container">$f: \Bbb R \to \Bbb R$</span> be twice differentiable. Then <span class="math-container">$$ \tag{**}\sup_{x \in \Bbb R} \left| f'\left( x\right) \right| ^{2}\le 4\sup_{x \in \Bbb R} \left| f\left( x\right) \right| \sup_{x \in \Bbb R} \left| f''\left( x\right) \right|$$</span></p>
</blockquote>
<p>which goes back to <a href="https://en.wikipedia.org/wiki/Edmund_Landau" rel="nofollow noreferrer">Edmund Landau</a>. See </p>
<ul>
<li><a href="https://en.wikipedia.org/wiki/Landau%E2%80%93Kolmogorov_inequality" rel="nofollow noreferrer">Landau–Kolmogorov inequality</a>,</li>
<li><a href="https://math.stackexchange.com/questions/257020/prove-sup-left-f-left-x-right-right-2-leqslant-4-sup-left-f-left">Prove $\sup \left| f'\left( x\right) \right| ^{2}\leqslant 4\sup \left| f\left( x\right) \right| \sup \left| f''\left( x\right) \right| $</a>,</li>
<li><a href="https://math.stackexchange.com/questions/38811/is-there-a-bounded-function-f-with-f-unbounded-and-f-bounded">Is there a bounded function $f$ with $f'$ unbounded and $f''$ bounded?</a>.</li>
</ul>
<p>The proofs resemble each other: We have
<span class="math-container">$$
f(t) = f(0) + t f'(0) + \int_{u=0}^t \int_{v=0}^u f''(v) \, dv
$$</span>
which implies
<span class="math-container">$$
0 \le \frac{t^2}2 \sup_{x \in \Bbb R} \left| f''\left( x\right) \right| - t f'(0) + 2 \sup_{x \in \Bbb R} \left| f\left( x\right) \right| \, .
$$</span>
Then <span class="math-container">$t$</span> is chosen such that the right-hand side is minimal. The same is done in above prove for sequences, only that <span class="math-container">$n$</span> is restricted to integers and cannot be chosen arbitrarily.</p>
<p>Landau also proved that the factor <span class="math-container">$4$</span> in <span class="math-container">$(**)$</span> is best possible. It would be interesting to know if <span class="math-container">$4$</span> is also the best possible factor for sequences in <span class="math-container">$(*)$</span>.</p>
|
122,945 | <p>Let $f:S^n\to C$ be a continuous function, $n\geq 1$. When $n=1$, this is a well-known theorem, called Kellog's theorem (or sometimes Kellog-Warschawski's theorem) which states the following</p>
<p>Theorem: Fix $k \geq 0, 0<\alpha<1$. Let $f\in C^{k,\alpha}(S^1)$. Then its harmonic extension $H(f)$, which is the solution to the Dirichlet problem on the unit disk $D$ with boundary value $f$, is in $C^{k, \alpha}(D)$.</p>
<p>My main question is: is the above true for $n\geq 2$ as well? Any refernces/ suggestions?</p>
<p>While I don't know exactly a complete reference for the proof, but I have read the following theorem mentioned in the book "Boundary Behaviour of Conformal maps" by Christian Pommerenke which states:</p>
<p>Let $F:D\to\Omega $ be a conformal homeomorphism of $D$ onto a Jordan domain $\Omega$ whose boundary curve $\partial\Omega$ has a $C^{k,\alpha}$ -parametrization. Then $f\in C^{k, \alpha}(D)$. Note that any conformal homeomorphism $F$ of $D$ onto a Jordan domain extends to the boundary of $D$, by Caratheodory's extension theorem.</p>
| Nagaraj Iyengar | 167,228 | <p>In respect of your main question, the answer is yes.
Please refer to the Algebraic Lemma on pp378 of the article
C^(1/,1/3)- regularity in the Dirichlet problem,
available at: <a href="https://www.sciencedirect.com/science/article/pii/S0898122107001927" rel="nofollow noreferrer">https://www.sciencedirect.com/science/article/pii/S0898122107001927</a></p>
|
3,811,753 | <p>Show that the equation:</p>
<p><span class="math-container">$$
y’ = \frac{2-xy^3}{3x^2y^2}
$$</span></p>
<p>Has an integration factor that depends on <span class="math-container">$x$</span> And solve it that way.</p>
<hr />
<p>Already we got to:</p>
<p><span class="math-container">$$
y’ + \frac{xy^3}{3x^2y^2} = \frac{2}{3x^2y^2}
$$</span></p>
<p>Therefore:</p>
<p><span class="math-container">$$
y’ + \frac{1}{3x}y = \frac{2}{3x^2}y^{-2}
$$</span></p>
<p>But, in order to get an integration factor, shouldn’t we have a linear equation? Of the form:</p>
<p><span class="math-container">$$
y‘ + p(x)y = g(x)
$$</span></p>
<p>That way getting the integration factor:</p>
<p><span class="math-container">$$
\mu = ke^{\int p(x)}, k \in R
$$</span></p>
<p>But what we have is a non-linear equation, so how could an integration factor exists?</p>
<p>Thanks.</p>
| user577215664 | 475,762 | <p><span class="math-container">$$y’ = \frac{2-xy^3}{3x^2y^2}$$</span>
Is linear if you substitute <span class="math-container">$w=y^3$</span> and <span class="math-container">$w'=3y^2y'$</span>
<span class="math-container">$$3y^2y’ = \frac{2-xy^3}{x^2}$$</span>
<span class="math-container">$$w' = \frac{2-xw}{x^2}$$</span>
<span class="math-container">$$xw'+w = \dfrac 2x$$</span>
<span class="math-container">$$(wx)'=\dfrac 2x$$</span>
Integrate.</p>
|
230,504 | <p>Again, this question is related (**) to a <a href="https://mathoverflow.net/questions/101700/large-cardinals-without-the-ambient-set-theory?rq=1">previous one</a>:</p>
<p>in standard books on basic set theory, after stating the axioms of ZFC, ordinal numbers are introduced early on. Afterwards cardinals appear: they are special ordinals which are minimal with respect to equinumerosity. </p>
<p>Ordinals and cardinals live happily within standard set theory.
Nothing wrong with that, but <strong>what about axiomatizing them <em>directly</em>, ie without the underlying set theory</strong>? </p>
<p>What I mean is: develop a first-order theory of some number system (the intended ordinals), such that they are totally ordered, there is an initial limit ordinal, and satisfy induction with respect to formulas in the language of ordinal arithmetic (here, of course, one will have to adjust the induction schema to accommodate the limit case).</p>
<p>The cardinals could be introduced by adding an order-preserving operator $K$ on the ordinal numbers mimicking their definition in ZFC: cardinals would then be the fixed points of $K$.</p>
<ol>
<li>has some direct axiomatization along these lines been fully developed? I would suspect that the answer is in the affirmative, but I have no refs. </li>
<li>the induction schema would be limited to first order formulae, so, assuming that the answer to 1 is yes, is there a theory of non-standard models of Ordinal Arithmetic? </li>
<li>Assuming 1 AND 2, what about weaker induction schemas for ordinals (*)? </li>
</ol>
<p>(*) I am thinking again of formal arithmetics and the various sub-systems of Peano</p>
<p>(**) it is not the same, though: here I am asking for a direct axiomatization of ordinals, and indirectly of cardinals, via the operator $k$</p>
| Mohammad Golshani | 11,115 | <p>I would like to suggest the theory of "<a href="https://books.google.com/books/about/Ordinal_algebras.html?id=cC44AAAAMAAJ">ordinal algebras</a>" and "<a href="https://books.google.com/books?id=9pM6AAAAMAAJ&q=inauthor:%22Alfred+Tarski%22+cardinal&dq=inauthor:%22Alfred+Tarski%22+cardinal&hl=en&sa=X&ved=0ahUKEwjxnvrSo-fKAhWHVxQKHdd4BWwQ6AEIHDAA">cardinal algebras</a>". There are books with the same title by Tarski.</p>
<p>Mathscinet review of the book cardinal algebras:</p>
<blockquote>
<p>This book is an axiomatic investigation of the novel types of algebraic systems which arise from three sources: the arithmetic of cardinal numbers; the formal properties of the direct product decompositions of algebraic systems; the algebraic aspects of invariant measures, regarded as functions on a field of sets. </p>
</blockquote>
<p>Mathscinet review of the book ordinal algebras:</p>
<blockquote>
<p>An ordinal algebra is an additively written, but not ordinarily commutative, associative system, equipped with a suitably axiomatized operation of simply infinite addition, $∑^∞_1a_ν$, and an operation of conversion, $a^∗$. (Infinite addition is characterized and employed "only insofar as [it is] involved in the study of finite addition''.) The additive theory of order types provides a familiar application, although far from exhausting the interest of the theory. The theory of ordinal algebras differs from that presented in the author's "Cardinal algebras'' principally in the non-commutativity of addition. </p>
</blockquote>
|
3,350,021 | <blockquote>
<p>We have the following quadratic equation:</p>
<p><span class="math-container">$2x^2-\sqrt{3}x-1=0$</span> with roots <span class="math-container">$x_1$</span> and <span class="math-container">$x_2$</span>.</p>
<p>I have to find <span class="math-container">$x_1^2+x_2^2$</span> and <span class="math-container">$|x_1-x_2|$</span>.</p>
</blockquote>
<p>First we have: <span class="math-container">$x_1+x_2=\dfrac{\sqrt{3}}{2}$</span> and <span class="math-container">$x_1x_2=-\dfrac{1}{2}$</span></p>
<p>So <span class="math-container">$x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=\dfrac{7}{4}$</span></p>
<p>Can someone help me with the second one?</p>
<p>I forgot to tell that solving the equation is not an option in my case.</p>
| Quanto | 686,284 | <p>Take the step below,</p>
<p><span class="math-container">$$(x_1-x_2)^2= x_1^2+x_2^2 - 2x_1x_2=\dfrac{7}{4}-2(-\frac 12) = \frac{11}{4}
$$</span></p>
<p>Thus, </p>
<p><span class="math-container">$$|x_1-x_2|=\frac{\sqrt{11}}{2}$$</span></p>
|
3,413,364 | <blockquote>
<p>Consider the set of points <span class="math-container">$$O = \{ x \in P \mid \alpha^* = C^T x \}$$</span> where <span class="math-container">$P \subseteq \mathbb R^n$</span> is a closed convex set, <span class="math-container">$C \in \mathbb R^n$</span> and <span class="math-container">$\alpha^* = \min \{ C^Tx \}$</span>. Then, <span class="math-container">$O$</span> is closed convex set.</p>
</blockquote>
<p>This seems a pretty simple statement in my linear programming class but I am unsure how to show it formally. I can easily show it is a convex set but I am not sure how to show it is a closed set.</p>
| gerw | 58,577 | <p>You have
<span class="math-container">$$ O = P \cap \{x \in \mathbb R^n | \alpha^* = C^\top x\},$$</span>
i.e., it is an intersection of two closed sets. Hence it is closed.</p>
|
227,109 | <p>I keep mixing them up, because they are very similar.</p>
<p>Some contrapositives resemble some contradictions.</p>
| ncmathsadist | 4,154 | <p>The contrapositive says that to argue $P\implies Q$, you instead argue $\sim Q\implies \sim P$.</p>
<p>Argument by contradiction is done by assuming $P$ and showing $P \implies
\rm{False}$. </p>
<p>Proving there is an infinity of primes is done by contradiction. You assume that there are finitely many. You take their product and add 1; you are forced to conclude this new object is relatively prime to every prime, which is impossible. The initial premise implies false, so it must be false.</p>
<p>Here is the contrapositive. Suppose that $ab$ odd. Then $a$ and $b$ is odd. let us argue by the contrapositive. Suppose $a$ or $b$ is even. Wlog, assume $a$ is even; write $a = 2n$ for an integer $n$. Then $ab = 2n = 2nb$; Since $2|ab$,
$ab$ is even.</p>
|
543,712 | <p>I am stuck on the following problem that says:</p>
<blockquote>
<p>Let <span class="math-container">$p,q$</span> be 2 complex numbers with <span class="math-container">$|p|<|q|$</span>. Let <span class="math-container">$$f(z)=\sum\{3p^n-5q^n\}z^n$$</span> Then the radius of convergence of <span class="math-container">$f(z)$</span> is :</p>
<ol>
<li><p><span class="math-container">$|q|$</span></p>
</li>
<li><p><span class="math-container">$|p|$</span></p>
</li>
<li><p>At least <span class="math-container">$\frac{1}{|q|}$</span></p>
</li>
<li><p>At most <span class="math-container">$\frac{1}{|q|}$</span></p>
</li>
</ol>
</blockquote>
<p><em>My Attempt</em>: <span class="math-container">$f(z)=\sum(3p^n-5q^n)z^n=\sum\{3p^n\}z^n-\sum\{5q^n\}z^n=3\sum(pz)^n-5\sum(qz)^n$</span>. Now for convergence,we must have <span class="math-container">$|pz|<1 \implies |z|<\frac{1}{|p|}$</span> and <span class="math-container">$|qz|<1 \implies |z|<\frac{1}{|q|}$</span>. Also we are given that <span class="math-container">$|p|<|q| \implies \frac{1}{|q|} <\frac{1}{|p|}$</span>.</p>
<p>Now,I am bit confused. Can someone help? Thanks in advance for your time.</p>
| angryavian | 43,949 | <p>Yes, there are $P(13,5)$ ways to choose five people to line up <em>in a certain order</em> to form the committee. But you count each committee $5!$ times (for example {Amy, Bob, Carl, Doug, Ed} is the same as {Ed, Doug, Carl, Bob, Amy}, but you counted them as being different). So we need to divide by $5!$. By the way, I think you have a typo: the answer is $C(13,5) = 13!/(8!5!)$.</p>
|
543,712 | <p>I am stuck on the following problem that says:</p>
<blockquote>
<p>Let <span class="math-container">$p,q$</span> be 2 complex numbers with <span class="math-container">$|p|<|q|$</span>. Let <span class="math-container">$$f(z)=\sum\{3p^n-5q^n\}z^n$$</span> Then the radius of convergence of <span class="math-container">$f(z)$</span> is :</p>
<ol>
<li><p><span class="math-container">$|q|$</span></p>
</li>
<li><p><span class="math-container">$|p|$</span></p>
</li>
<li><p>At least <span class="math-container">$\frac{1}{|q|}$</span></p>
</li>
<li><p>At most <span class="math-container">$\frac{1}{|q|}$</span></p>
</li>
</ol>
</blockquote>
<p><em>My Attempt</em>: <span class="math-container">$f(z)=\sum(3p^n-5q^n)z^n=\sum\{3p^n\}z^n-\sum\{5q^n\}z^n=3\sum(pz)^n-5\sum(qz)^n$</span>. Now for convergence,we must have <span class="math-container">$|pz|<1 \implies |z|<\frac{1}{|p|}$</span> and <span class="math-container">$|qz|<1 \implies |z|<\frac{1}{|q|}$</span>. Also we are given that <span class="math-container">$|p|<|q| \implies \frac{1}{|q|} <\frac{1}{|p|}$</span>.</p>
<p>Now,I am bit confused. Can someone help? Thanks in advance for your time.</p>
| Arcane | 160,050 | <p><a href="https://math.stackexchange.com/questions/567606/when-to-use-permutations-or-combinations/849150#849150">When to use Permutations or Combinations</a></p>
<p>I've also illustrated the concept here. The basic difference is to understabd you have to just choose a community (i.e. selection) and not prepare a Seating chart for the selected committee (ordering)</p>
<p>Thus in this case the answer is C(13,5) is the way to select a committee of 5 people, while if the question was to prepare number of ways to seat 5 people in from 13 the answer would be P(13,5) or 5!xC(13,5).</p>
|
187,975 | <p>Let $\mu$ be a finite nonatomic measure on a measurable space $(X,\Sigma)$, and for simplicity assume that $\mu(X) = 1$. There is a well-known "intermediate value theorem" of Sierpiński that states that for every $t \in [0,1]$, there exists a set $S \in \Sigma$ with $\mu(S) = t$.</p>
<p>I would like to use the following stronger conclusion for such a measure: </p>
<blockquote>
<p>There exists a chain of sets $\{S_t \mid t \in [0,1]\}$ in $\Sigma$,
with $S_t \subseteq S_r$ whenever $0 \leq s \leq r \leq 1$, such that
$\mu(S_t) = t$ for all $t \in [0,1]$.</p>
</blockquote>
<p>(One can view this as the existence a right inverse to the map $\mu \colon \Sigma \to [0,1]$ in the category of partially ordered sets.)</p>
<p>This statement appears (albeit hidden within a proof) on the Wikipedia page for "<a href="http://en.wikipedia.org/wiki/Atom_%28measure_theory%29#Non-atomic_measures" rel="noreferrer">Atom (measure theory)</a>," and even includes a sketch for the proof! However, I would like to see some mention of this in the literature. I've checked the Wiki references and they both seem to prove the weaker statement. I looked in Fremiln's <em>Measure Theory</em>, vol. 2, and again found the weaker version but not the stronger. </p>
<p><strong>Question:</strong> Can anyone provide me with such a reference?</p>
<hr>
<p><strong>A proof.</strong> In case anyone stumbles to this page and wants to see a proof, I'll sketch one that is more constructive than the one that I linked to above. Set $S_0 = \varnothing$ and $S_1 = X$. By Sierpiński, there exists $S_{1/2} \in \Sigma$ of measure $1/2$. For each Dyadic rational $q = m/2^n \in [0,1]$ ($1 \leq m \leq 2^n$), we may proceed by induction on $n$ to construct each $S_q$. Now given $r \in [0,1]$, set $S_r = \bigcup_{q \leq r} S_q$. (This is essentially the same method of proof as the one in the reference provided in Ramiro de la Vega's answer.)</p>
| Salvo Tringali | 16,537 | <p>It is also a special case of Theorem 15 (p. 43) in:</p>
<blockquote>
<p>A. Fryszkowski, <em>Fixed Point Theory for Decomposable Sets</em>, Topological Fixed Point Theory and Its Applications <strong>2</strong>, Dordrecht: Kluwer Academic Publishers, 2004.</p>
</blockquote>
|
331,962 | <p>We have an first order ODE : </p>
<p>Equation1 : $y' + y = x$ ?
We can view the left-hand side as an operator acting on $y$. </p>
<p>In that case $L=(d/dx + 1)$ </p>
<p>$L(y_1) = x$<br>
$L(y_2)=x$<br>
$L(y_1+y_2)=x$<br>
So, clearly $L(y_1+y_2) = x \neq L(y_1)+L(y_2) = 2x$ </p>
<p>So why is $y'+y=x$ is a linear ODE ?</p>
| nerdy | 66,708 | <p>Try to solve my confusion, please. </p>
<p>L[y1] = y1' + y1 </p>
<p>L[y2] = y2' + y2 </p>
<p>In this case :<br>
L[y1+y2] = (y1+y2)' + y1 + y2 = L[y1] + L[y2] </p>
<p>At the same time, by the ODE :<br>
L[y] = x </p>
<p>So, L[y1]=L[y2]=x<br>
L[y1+y2] = x as well </p>
<p>So, L[y1]+L[y2] =/= L[y1+y2] </p>
<p>Where is the error on my logic that finds both L[y1]+L[y2] =/= L[y1+y2] and L[y1] + L[y2] = L[y1+y2] ?</p>
|
3,294,123 | <p>Define <span class="math-container">$f(x)=x^{-1}(\log x)^{-2}$</span> if <span class="math-container">$0<x<\frac{1}{2}$</span>, <span class="math-container">$f(x)=0$</span> on the rest of <span class="math-container">$R$</span>. Then <span class="math-container">$f \in L^1(R)$</span>. Show that
<span class="math-container">$$(Mf)(x) \ge |2x \log(2x)|^{-1} \;\;(0<x<1/4)$$</span> so that <span class="math-container">$\int_0^1 (Mf)(x)dx=\infty.$</span></p>
<p><span class="math-container">$$
(Mf)(x) = \sup_{0<r<\infty} \frac{1}{m(B_r)}\int_{B_r(x)}|f(y)|\mathrm{d}y
$$</span>
where <span class="math-container">$B_r(z)$</span> is a ball centered around <span class="math-container">$x$</span> with radius <span class="math-container">$r$</span>. </p>
<p>I am stuck with proving this lower bound for the maximality function. I would greatly appreciate any help.</p>
| Matematleta | 138,929 | <p>The other answer is the slick way to go, but the hint reduces the problem to a calculation:</p>
<p>Take <span class="math-container">$0<x<1/4$</span> and <span class="math-container">$r=x$</span>. Then, </p>
<p><span class="math-container">$Mf(x)\ge \frac{1}{B_r(x)}\int_{B_r(x)} |f|=\frac{1}{2x}\int_{0}^{2x}|f(t)|dt=$</span></p>
<p><span class="math-container">$\frac{1}{2x}\cdot \underset{\delta\to 0^+ } \lim \int_{\delta}^{2x}|f(t)|dt=\frac{1}{2x}\cdot \underset{\delta\to 0^+ } \lim \left |\frac{1}{\ln 2x}-\frac{1}{\ln \delta}\right |=\frac{1}{2x\ln 2x}$</span></p>
<p>from which the result follows. </p>
|
43,956 | <p>There is this example at the Wikipedia article on Quotient spaces (QS):</p>
<blockquote>
<p>Consider the set $X = \mathbb{R}$ of all real numbers with the ordinary topology, and write $x \sim y$ if and only if $x−y$ is an integer. Then the quotient space $X/\sim$ is homeomorphic to the unit circle $S^1$ via the homeomorphism which sends the equivalence class of $x$ to $\exp(2πix)$.</p>
</blockquote>
<p>I understand relations, equivalence relation and equivalence class but quotient space is still too abstract for me. This seems like a simple enough example to begin with.</p>
<p>I understand (sort of) the definition but I can't visualize. And by this example and others, there is a lot of visualizing going on here! torus, circles etc.</p>
| C-star-W-star | 79,762 | <p>The quotient topology is tricky! ...</p>
<p><strong>There's a number of aspects to consider; all of them basically boil down to:</strong></p>
<ol>
<li>Equip the quotient topology: $\pi:(X,\mathcal{S})\to Y$</li>
<li>Check for quotient map: $\pi:(X,\mathcal{S})\to (Y,\mathcal{T})$</li>
</ol>
<p><strong>1.) Why do we take the quotient topology?</strong></p>
<p>Given a plain space one wants to give it a topology that is good enough in the sense:
$g\text{ continuous}\iff g\circ\pi\text{ continuous}$<br>
In other words, we don't want to artificially create new continuous maps plus we don't want to accidentally loose continuous maps.<br>
Luckily, this can be guaranteed by imposing the quotient topology!</p>
<p>Given a plain space one wants to give it a topology that is good enough in the sense:
$f\text{ continuous}\iff\pi\circ f\text{ continuous}$<br>
So one would like to maintain continuous maps with our plain space as codomain rather than as domain.<br>
Unfortunately, this cannot be guaranteed in general by specific choice of some topology!</p>
<p>So that is all we can get!<br>
(<em>By the way, this is the same reason for the subspace topology as well as the product topology</em>)</p>
<p><strong>2.) How can we check and compare spaces for quotient topology?</strong></p>
<p>a) Checking for quotient map:</p>
<p>Now, we are given already some topology. Is it the quotient topology?<br>
Obviously, the projection must be necessarily continuous and surjective in order to be a quotient map, that is the given topology is the quotient topology.<br>
However, this is not enough!!! (i.e. $\pi:[0,1)\to\mathbb{S}^1: s\mapsto \mathrm{e}^{2\pi\mathrm{i}s}$)<br>
There's actually a number of lemmata providing sufficient conditions for it to be a quotient map, among them the one on saturated sets or the one on closed sets.</p>
<p>b) Comparing quotient maps:</p>
<p>Assume you are given now two topological spaces together with their own quotient maps and they seem to describe basically the same object, that is they seem to be isomorphic:
$\pi_1:(X,\mathcal{S})\to(Y_1,\mathcal{T}_1)$ and $\pi_2:(X,\mathcal{S})\to(Y_2,\mathcal{T}_2)$<br>
Then, in order to prove that they are basically the same it suffices to check wether they make the same identifications:
$\pi_1(x)=\pi_1(\tilde{x})\iff\pi_2(x)=\pi_2(\tilde{x})$<br>
(<em>That is quite neat since one does not have to check topological properties</em>)</p>
<p><strong>2.)* How can we practically work with these?</strong></p>
<p>There's a central result called the 'closed map lemma' that offers an astonishingly simple way to prove strong statements:</p>
<ul>
<li>Quotient map?</li>
<li>Topological embedding?</li>
<li>Homeomorphism?</li>
</ul>
<p>All one has to check is continuity!!! (plus some little things)</p>
<p>In particular, this is quite useful in your case to prove the realization of the unit sphere as a quotient space. (<em>Both of the preceding answers by Henno and by Aaron made use of this!</em>)<br>
Here's a draft how this goes:<br>
$\pi:[0,1]\to[0,1]/\sim:s\mapsto [s]$ and $\epsilon:[0,1]\to\mathbb{S}^1:s\mapsto\mathrm{e}^{2\pi\mathrm{i}s}$<br>
$\epsilon\text{ continuous and surjective}\Rightarrow\epsilon\text{ quotient map}$ ...with $[0,1]\text{ compact}$ and $\mathbb{S}^1\text{ Hausdorff}$<br>
$\pi(s)=\pi(\tilde{s})\text{ iff }\epsilon(s)=\epsilon(\tilde{s})\Rightarrow[0,1]/\sim\cong\mathbb{S}^1$</p>
<p>... There's a lot more to discuss on quotient spaces but that should give you some first insights.</p>
|
742,216 | <p>$a$, $b$, $c$, $d$ are rational numbers and all $> 0$.</p>
<p>$\max \left\{\dfrac{a}{b} , \dfrac{c}{d}\right\} \geq \dfrac{a+c}{b+d}\geq \min
\left\{\dfrac{a}{b} , \dfrac{c}{d}\right\}$</p>
<p>Hope someone can help me with this one. How would you go on proving the validity? Thanks in advance.</p>
| Martin Sleziak | 8,297 | <p>Let $m=\min\{a/b,c/d\}$. This means that
$$ m \le \frac ab \qquad \text{and} \qquad m\le \frac cd,$$
which is equivalent to $bm\le a$ and $dm\le c$. Thus we get
$$a+c\ge bm+dm=(b+d)m$$
which is equivalent to
$$\frac{a+c}{b+d}\ge m.$$</p>
<p>The proof of the inequality for maximum is similar.</p>
<p>(Note that in all steps, where we multiplied the inequality by some number, this number was positive. Therefore the sing of the inequality was not changed.) </p>
|
55,404 | <p>I have been searching for a version of the isoperimetric inequality which is something like:</p>
<p>$P(\Omega) - 2\sqrt{\pi} Vol(\Omega)^{1/2} \geq \pi (r_{out}^2 - r_{in}^2)$ where $r_{out}$ and $r_{in}$ are the inner and outer radius of a given set. There are of course details which I am missing such as what kind of sets this applies to (clearly connected and possibly simply connected). I was hoping somebody may recognize this inequality and be able to direct me to a source for it along with a proof.</p>
<p><strong>Update:</strong> I'm curious if anyone can direct me to a some papers which relate the isoperimetric deficit to Hausdorff distance. Such as:
$P(\Omega)^2 - 4\pi Vol(\Omega) \geq C d_H(\Omega,B)^2$ whre $B$ is a sphere in $\mathbb{R}^2$
which may be the inner or outer circle.</p>
<p><strong>Update April 12:</strong> I would like to know if the first Bonnesen inequality written below is strictly stronger than the one in higher dimensions? In particular, if one considers the Fraenkel assymetry $\alpha(\Omega) = \min_B |\Omega \Delta B|$ where $|B|=|\Omega|$, does it hold on a bounded domain that</p>
<p>$ r_{out}^2 - r_{in}^2 \leq C \alpha(\Omega)$,</p>
<p>for some constant $C>0$? This seems like it should be true but I can't seem to find a concise proof of it.</p>
| Mark Meckes | 1,044 | <p>A classical result along these lines is <a href="http://en.wikipedia.org/wiki/Bonnesen%27s_inequality" rel="nofollow">Bonnesen's inequality</a>, which states
$$
L^2 - 4\pi A \ge \pi^2 (r_{out} - r_{in})^2,
$$
where $L$ is the length and $A$ is the enclosed area of a simple planar closed curve. There are many other results along these lines, called "stability estimates" for the isoperimetric inequality. There are several pointers to the literature in Note 6 following section 6.2 of Schneider's book <em>Convex Bodies: the Brunn-Minkowski Theory</em>.</p>
<p><strong>Added:</strong> Bonnesen's inequality also suffices for the updated question. If $B_{in}$ and $B_{out}$ are the inner and outer disks, respectively, then since $B_{in} \subseteq \Omega \subseteq B_{out}$,
$$
d_H(\Omega, B) \le d_H(B_{in},B_{out}) \le 2r_{out} - 2r_{in}
$$
(the extreme case in the latter inequality being when the circles are tangent), so you get your desired result with $C = \pi^2/4$.</p>
|
670,813 | <p>Let $Y$ be a closed subspace of a compact space $X$. Let $i:Y \to X$ the inclusion and $r:X \to Y$ a retraction ($r \circ i = Id_Y$). I have to prove that exists this short exact sequence
$$ 0 \to K(X,Y) \to K(X) \to K(Y) \to 0.$$
Then I have to verify that $K(X) \simeq K(X,Y) \oplus K(Y)$. How can I do it? I think that $K(X,Y) = \tilde{K}(X/Y).$ Thank you very much.</p>
| Georges Elencwajg | 3,217 | <p>This is purely formal and relies on the fact that $K$ is a contravariant functor from topological spaces to abelian groups (actually to commutative rings but this is not needed here).<br>
Since $r\circ i=Id_Y$ we get $i^*\circ r^*=Id_{K(Y)}$ so that $r^*:K(Y)\to K(X)$ is a section of $i^*:K(X)\to K(Y)$ and your exact sequence of abelian groups $ 0 \to K(X,Y) \to K(X) \stackrel {i^*}{\to} K(Y) \to 0$ splits , yielding the required isomorphism $K(X) \simeq K(X,Y) \oplus K(Y)$. </p>
<p>By the way, it is indeed true that $K(X,Y) = \tilde{K}(X/Y)$ but this fact is irrelevant to the question at hand.</p>
|
3,288,010 | <p>The following snippet is from Adamek, Rosicky:Algebra and local presentability,how algebraic are.</p>
<p>It is unclear to me the end of Example 5.1:</p>
<p>Since <span class="math-container">$e$</span> is the coequalizer of <span class="math-container">$\bar{u}_1,\bar{u}_2$</span> in <span class="math-container">$\mathbf{Pos}$</span>, we conclude that <span class="math-container">$W$</span> does not preserve <span class="math-container">$W-$</span>split coequalizers.</p>
<p>Why?</p>
<p>The snippet:</p>
<p><a href="https://i.stack.imgur.com/ElaYF.jpg" rel="nofollow noreferrer"><img src="https://i.stack.imgur.com/ElaYF.jpg" alt="enter image description here"></a></p>
| lhf | 589 | <p>Another approach is simply to compute <span class="math-container">$x^2+2y^2 \bmod 8$</span> for <span class="math-container">$x,y =0, \dots 7$</span>. The result is <span class="math-container">$0, 1, 2, 3, 4, 6$</span>. Therefore, if <span class="math-container">$z \equiv 5,7 \bmod 8$</span>, then <span class="math-container">$z$</span> is not of the form <span class="math-container">$x^2+2y^2$</span>. This has nothing to do with <span class="math-container">$z$</span> being prime or not.</p>
|
7,575 | <p>How could I display text that flashed red for a half second or so and then reverted to black? (Or was put in bold and reverted to normal, etc.)</p>
| Szabolcs | 12 | <p>It's worth pointing out that several of the other answers keep dynamically re-evaluating an expression until infinity.</p>
<p>Here's a solution that fades out the text gradually, and more importantly: it does not keep re-evaluating the <code>Dynamic</code> expression until infinity. The third argument of <code>Clock</code> (and the fact that <code>Dynamic</code> is smart when handling <code>Clock</code>) can be used to terminate the evaluation.</p>
<pre><code>flash[expr_] :=
Dynamic[Style[expr, Blend[{Red, Black}, Clock[{0, 1}, 1, 1]]]]
flash["Boo!"]
</code></pre>
<p>(Update: I overlooked that István's original answer uses <code>Clock</code> in the same way.)</p>
|
7,575 | <p>How could I display text that flashed red for a half second or so and then reverted to black? (Or was put in bold and reverted to normal, etc.)</p>
| Alexei Boulbitch | 788 | <p>As a variation of answer to this question, here is an approach I used for lectures. Here are the functions Ac and Pl that either accentuate or leave the text plain:</p>
<pre><code>Ac[expr_] :=
DynamicModule[{c1 = 0}, EventHandler[
Dynamic[
If[c1 == 0,
Style[expr, Black, Plain, 22, Italic] // ExpressionCell,
Dynamic @ If[Clock[1, 0.7, 2] < .5,
Style[expr, Gray, Plain, 22, Italic] // ExpressionCell,
Style[expr, RGBColor[0.8, 0, 0], Bold, 22, Italic]] // ExpressionCell]],
{"MouseDown" :> (c1 = c1 /. {0 -> 1, 1 -> 0})}
]];
</code></pre>
<p>and </p>
<pre><code>Pl[expr_] := ExpressionCell[Style[expr, Black, Plain, 22, Italic]];
</code></pre>
<p>And here is an example of its application:</p>
<pre><code>F = Row[{"AA" // Pl, "+BB" // Ac, "+CC" // Pl, "+DD" // Ac}]
</code></pre>
<p>The strings "+BB" and "+DD" are accentuated. The accentuation is shown upon the mouseclick on the corresponding elements. </p>
|
20,567 | <p>Excuse me if the language is a bit off I'm not a native English speaker. I've been studying(self study) computers and programming for a little over two years, but do not have much education. I've been taking a math class for the last 11 weeks and start an new one next week. I've been using Unix/Linux and Stackoverflow on this network for this time and am pretty terrified at the level the math is in here. I've been browsing questions and can't find any for my level. I'm pretty serious in learning math and have books on Basic Maths, Algebra I, Algebra II, Linear Algebra, Pre-Calculus and Calculus and plan on getting more.</p>
<p>I have sometimes gotten stuck in my learning and Unix/Linux has helped me a lot in the computers. </p>
<p>Can I ask questions here about things where I get stuck at or don't know how to move past a point in my math, at my level? I haven't found anything which says otherwise and also it say's all levels in the help.</p>
<p>The topics of the course I just finished where: Addition/Subtraction, Multiplication, Fractions, Algebra, Equations, Power and Square-root. The course I start at next week is the next step, so pretty much I just learned the basics and I'm starting my math career.</p>
| mrf | 19,440 | <p>Questions at every level are welcome.</p>
<p>Just make sure to provide enough detail and context to your question, preferably pinpointing what you are having problems with. Don't just copy-paste textbook assignment.</p>
<p>Have a look, for example at the questions tagged <a href="https://math.stackexchange.com/questions/tagged/arithmetic" class="post-tag" title="show questions tagged 'arithmetic'" rel="tag">arithmetic</a> to get a feeling for what questions are well received, and what are not.</p>
|
4,008,152 | <p>Question itself: Throw a coin one million times. What is the expected number of sequences of six tails, if we <strong>do not allow overlap</strong>?</p>
<p>I know when overlap is allowed, the answer is (1,000,000-5)/(2^6). Not sure if we can just do (1,000,000-5)/(2^6) divided by 6 if overlap is not allowed?</p>
<p>Some clarifications:</p>
<p>For example, if part of the sequence is "one H, nine T, then one H", we would count 1 sequence of six tails. (When overlap is allowed, we can count three times because each of the first 3 T can start a sequence of six tails; However, this question does not allow overlap, so 9T can only be counted as containing <strong>one</strong> sequence of six tails)</p>
<p>If part of the sequence is "one H, thirteen T, then one H", we would count 2 sequences of six tails.</p>
| user334639 | 221,027 | <p>I think I can compute that with an error of plus or minus 1.</p>
<p>This is a sketchy argument that you can make rigorous using Ergodic Theory or Palm measures.</p>
<p>Let us group runs of T from left to right, so a run of 14 T's has a run of 6 starting at position 1, another run starting at position 7, and two single T's at positions 13 and 14 that are not grouped.</p>
<p>Imagine a doubly-infinite sequence of H and T</p>
<p>You can construct it as a doubly infinite sequence of H and an doubly infinite iid sequence of H^k where k+1 is random Geometric(0.5). The average distance between H's is 2 and the average number of non-overlapping runs TTTTTT before the next H equals 1/63.</p>
<p>So a proportion of 1/126 of the integers will be the leftmost point of a non-overlapping group of 6.</p>
<p>The answer would be 1,000,000/126, except that this is not counting the case where a sequence of T's start at some x<0 and ends at 0<x<6 with more T's in x+1,...,6, and it counts the case of an x near the end which is the start of a run that actually ends after the interval. So 1,000,000/126 is the expectation of a random variable that can differ from your random variable by at most 1 unit.</p>
<p>You can probably improve it to an exact number but I'm quite happy with this approximation.</p>
|
4,008,152 | <p>Question itself: Throw a coin one million times. What is the expected number of sequences of six tails, if we <strong>do not allow overlap</strong>?</p>
<p>I know when overlap is allowed, the answer is (1,000,000-5)/(2^6). Not sure if we can just do (1,000,000-5)/(2^6) divided by 6 if overlap is not allowed?</p>
<p>Some clarifications:</p>
<p>For example, if part of the sequence is "one H, nine T, then one H", we would count 1 sequence of six tails. (When overlap is allowed, we can count three times because each of the first 3 T can start a sequence of six tails; However, this question does not allow overlap, so 9T can only be counted as containing <strong>one</strong> sequence of six tails)</p>
<p>If part of the sequence is "one H, thirteen T, then one H", we would count 2 sequences of six tails.</p>
| Sal Elder | 1,025,101 | <p>We can break this into two problems. First, what's the expected number of sequences of <span class="math-container">$T^{6k}$</span> without a preceding <span class="math-container">$H$</span>? (I.e., number of <span class="math-container">$HT^{6k}$</span> in any position, or <span class="math-container">$T^{6k}$</span> at the beginning.) Call it <span class="math-container">$E_k.$</span> Second, sum over <span class="math-container">$kE_k$</span> to get the final answer.</p>
<p>We have <span class="math-container">$10^6-6k$</span> windows of length <span class="math-container">$6k+1.$</span> The probability that a given window contains <span class="math-container">$HT^{6k}$</span> is <span class="math-container">$1/2^{6k+1},$</span> and the probability that the beginning of the sequence is <span class="math-container">$T^{6k}$</span> is <span class="math-container">$1/2^{6k}.$</span> So <span class="math-container">$$E_k=\frac{10^6-6k+2}{2^{6k+1}}.$$</span></p>
<p>Next, <span class="math-container">$k$</span> ranges from 1 to 166,666. So the answer is
<span class="math-container">$$\sum_{k=1}^{166,666}kE_k\approx 8062.45.$$</span>
I used Wolfram Alpha for the sum, which is probably not allowed in an interview.</p>
|
4,008,152 | <p>Question itself: Throw a coin one million times. What is the expected number of sequences of six tails, if we <strong>do not allow overlap</strong>?</p>
<p>I know when overlap is allowed, the answer is (1,000,000-5)/(2^6). Not sure if we can just do (1,000,000-5)/(2^6) divided by 6 if overlap is not allowed?</p>
<p>Some clarifications:</p>
<p>For example, if part of the sequence is "one H, nine T, then one H", we would count 1 sequence of six tails. (When overlap is allowed, we can count three times because each of the first 3 T can start a sequence of six tails; However, this question does not allow overlap, so 9T can only be counted as containing <strong>one</strong> sequence of six tails)</p>
<p>If part of the sequence is "one H, thirteen T, then one H", we would count 2 sequences of six tails.</p>
| user1122507 | 1,122,507 | <p>I was inspired by Sal Elder, but I think there are some problems with his answer.
Below is my answer:</p>
<p>First step: calculate <span class="math-container">$E_k$</span>:the expected number of sequences of more than 6k' T with a preceding H(or at the begining).</p>
<p>Second step: sum over <span class="math-container">$k(E_k-E_{k+1})$</span>, where <span class="math-container">$E_k-E_{k+1}$</span> is happen to be the expected number of sequences whose T is more than 6k but less than 6(k+1).</p>
<p>In this way, we count each sequence exactly once without overlapping.</p>
|
18 | <p>Some teachers make memorizing formulas, definitions and others things obligatory, and forbid "aids" in any form during tests and exams. Other allow for writing down more complicated expressions, sometimes anything on paper (books, tables, solutions to previously solved problems) and in yet another setting students are expected to take questions home, study the problems in any way they want and then submit solutions a few days later.</p>
<p>Naturally, the memory-oriented problem sets are relatively easier (modulo time limit), encourage less understanding and more proficiency (in the sense that the student has to be efficient in his approach). As the mathematics is in big part thinking, I think that it is beneficial to students to let them focus on problem solving rather than recalling and calculating (i.e. designing a solution rather than modifying a known one). There is a huge difference between work in time-constrained environment (e.g. medical teams, lawyers during trials, etc.) where the cost of "external knowledge" is much higher and good memory is essential.
However, math is, in general (things like high-frequency trading are only a small part math-related professions), slow.</p>
<p>On the other hand, memory-oriented teaching is far from being a relic of the past. Why is this so? As this is a broad topic, I will make it more specific:</p>
<p><strong>What are the advantages of memory-oriented teaching?</strong></p>
<p><strong>What are the disadvantages of allowing aids during tests/exams?</strong></p>
| Community | -1 | <p>In many cases, there is a choice between memorizing and understanding. E.g., a calc student can memorize that <span class="math-container">$\cos'=-\sin$</span>, or they can understand that it's a phase shift of 90 degrees and figure out the direction of the phase shift by visualization. Understanding is better, but it requires three things: (1) a teacher who has this type of understanding, (2) a teacher who emphasizes and aids in this kind of understanding, (3) a student who is intellectually capable of it, and (4) a student who is willing to do it. An advantage of memory-oriented teaching may be that it works in cases where <span class="math-container">$\neg$</span>(1 and 2 and 3 and 4).</p>
<p>Especially in K-12 and community college education, there are many cases where 1 is false. When you get teachers who lack this level of understanding, they don't believe that it's useful or can substitute for a certain amount of memorization. For example, when I was a kid I did <span class="math-container">$6\times8$</span> by remembering <span class="math-container">$3\times8=24$</span> and then doubling that. But I have spoken to a lot of kids who were specifically told never to do this kind of thing, because it would be too slow. If they scribbled this kind of thing in the margin of their times-table test, they were penalized. This is what you get from teachers who don't themselves have the habit of understanding. They have never had the experience of doing this kind of memorization-avoidance strategy successfully, so they have never had the experience of seeing how it becomes faster and faster, until eventually it's very much like memorization.</p>
|
743,227 | <p>I've this question:</p>
<blockquote>
<p>Find the area of the intersection between the sphere $x^2 + y^2 + z^2 = 1$ and the cylinder $x^2 + y^2 - y = 0$.</p>
</blockquote>
<p>Is this second equation even a closed shape? If one were to plot points satisfying that equation, one gets things like $(2, \sqrt{-2})$, $(3, \sqrt{-6})$ and all that.</p>
<p><strong>Edit:</strong> I understand the equation for a circle and such, and have (with the help of everyone who answered) found my issue.</p>
<p>I was plugging in (whole) numbers that weren't in the codomain of the cylinder, similar to having, say, the equation of a circle $x^2 + y^2 = 16$ and plugging in $25$ for $x$—you will get a complex number for $y$. If one plugs in only numbers not in the domain/codomain, then the equation will not <em>seem</em> like the shape it should be. </p>
<p>Sorry for my shortsightedness, and thanks everyone for replying so promptly. :)</p>
| Mark Bennet | 2,906 | <p>The equation $x^2+y^2-y=0$ can be rewritten $x^2+(y-\frac 12)^2=\cfrac 14$.</p>
<p>For any value of $z$ this is a circle, so you should be able to see how this makes the figure a cylinder (like a straight line, a cylinder in this terminology has no ends).</p>
|
98,700 | <blockquote>
<p>Suppose you wanted to write the number 100000. If you type it in ASCII, this would take 6 characters (which is 6 bytes). However, if you represent it as unsigned binary, you can write it out using 4 bytes.</p>
</blockquote>
<p>(from <a href="http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/BitOp/asciiBin.html" rel="nofollow">http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/BitOp/asciiBin.html</a>)</p>
<p>My question: $\log_2 100,000 \approx 17$. So that means I need <code>17</code> bits to represent <code>100,000</code> in binary, which requires 3 bytes. So why does it say 4 bytes?</p>
| JRN | 18,398 | <p>This is more of a computer science/engineering question than a math question.</p>
<p>Look at <a href="http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Data/unsigned.html" rel="nofollow">http://www.cs.umd.edu/class/sum2003/cmsc311/Notes/Data/unsigned.html</a>. It asks you to "assume that a typical unsigned int uses 32 bits of memory." Programming languages and processors usually use an even number of bytes to represent data.</p>
|
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